6 Bézout theorem and a group law on a plane cubic curve 45

7 Morphisms of projective algebraic varieties 57

8 Quasi-projective algebraic sets 69

9 The image of a projective algebraic set 77

10 Finite regular maps 83

12 Lines on hypersurfaces 105

13 Tangent space 117

14 Local parameters 131

15 Projective embeddings 147

iiiiv CONTENTS

16 Blowing up and resolution of singularities 159

17 Riemann-Roch Theorem 175

Index 191Lecture 1

Systems of algebraic equations

The main objects of study in algebraic geometry are systems of algebraic equa-tions and their sets of solutions. Let k be a field and k[T1 , . . . , Tn ] = k[T ] bethe algebra of polynomials in n variables over k. A system of algebraic equationsover k is an expression {F = 0}F ∈S ,where S is a subset of k[T ]. We shall often identify it with the subset S. Let K be a field extension of k. A solution of S in K is a vector (x1 , . . . , xn ) ∈ nK such that, for all F ∈ S,

F (x1 , . . . , xn ) = 0.

Let Sol(S; K) denote the set of solutions of S in K. Letting K vary, we get

different sets of solutions, each a subset of K n . For example, let

S = {F (T1 , T2 ) = 0}

be a system consisting of one equation in two variables. Then

Sol(S; Q) is a subset of Q2 and its study belongs to number theory. Forexample one of the most beautiful results of the theory is the Mordell Theorem(until very recently the Mordell Conjecture) which gives conditions for finitenessof the set Sol(S; Q). Sol(S; R) is a subset of R2 studied in topology and analysis. It is a union ofa finite set and an algebraic curve, or the whole R2 , or empty. Sol(S; C) is a Riemann surface or its degeneration studied in complex analysisand topology.

12 LECTURE 1. SYSTEMS OF ALGEBRAIC EQUATIONS

All these sets are different incarnations of the same object, an affine algebraicvariety over k studied in algebraic geometry. One can generalize the notion ofa solution of a system of equations by allowing K to be any commutative k-algebra. Recall that this means that K is a commutative unitary ring equippedwith a structure of vector space over k so that the multiplication law in K is abilinear map K × K → K. The map k → K defined by sending a ∈ k to a · 1is an isomorphism from k to a subfield of K isomorphic to k so we can and wewill identify k with a subfield of K. The solution sets Sol(S; K) are related to each other in the following way.Let φ : K → L be a homomorphism of k-algebras, i.e a homomorphism of ringswhich is identical on k. We can extend it to the homomorphism of the directproducts φ⊕n : K n → Ln . Then we obtain for any a = (a1 , . . . , an ) ∈ Sol(S; K),

φ⊕n (a) := (φ(a1 ), . . . , φ(an )) ∈ Sol(S; L).

This immediately follows from the definition of a homomorphism of k-algebras

K 7→ Sol(S; K), φ → sol(S; φ)

equivalent if Sol(S; K) = Sol(S 0 , K) for any k-algebra K. An equivalence classis called an affine algebraic variety over k (or an affine algebraic k-variety). If Xdenotes an affine algebraic k-variety containing a system of algebraic equationsS, then, for any k-algebra K, the set X(K) = Sol(S; K) is well-defined. It iscalled the set of K-points of X. 3

braic variety denoted by Ank . It is called the affine n-space over k. We have, forany k-algebra K, Sol({0}; K) = K n .2. The system 1 = 0 defines the empty affine algebraic variety over k and isdenoted by ∅k . We have, for any K-algebra K,

∅k (K) = ∅.

We shall often use the following interpretation of a solution a = (a1 , . . . , an ) ∈

a ∈ Sol(S; K) ⇐⇒ eva (S) = {0}.

In particular, eva factors through the factor ring k[T ]/(S), where (S) stands forthe ideal generated by the set S, and defines a homomorphism of k-algebras

evS,a : k[T ]/(S) → K.

Conversely any homomorphism k[T ]/(S) → K composed with the canonical

surjection k[T ] → k[T ]/(S) defines a homomorphism k[T ] → K. The imagesai of the variables Ti define a solution (a1 , . . . , an ) of S since for any F ∈ S theimage F (a) of F must be equal to zero. Thus we have a natural bijection

Sol(S; K) ←→ Homk (k[T ]/(S), K).

It follows from the previous interpretations of solutions that S and (S) definethe same affine algebraic variety. The next result gives a simple criterion when two different systems of algebraicequations define the same affine algebraic variety.

same affine algebraic variety if and only if the ideals (S) and (S 0 ) coincide.

Proof. The part ‘if’ is obvious. Indeed, if (S) = (S 0 ), then for every F ∈ Swe can express F (T ) as a linear combination of the polynomials G ∈ S 0 withcoefficients in k[T ]. This shows that Sol(S 0 ; K) ⊂ Sol(S; K). The oppositeinclusion is proven similarly. To prove the part ‘only if’ we use the bijection4 LECTURE 1. SYSTEMS OF ALGEBRAIC EQUATIONS

from the Proposition 1.2 that S and S 0 define different algebraic varieties X andY . For every k-algebra K the set Sol(S; K) consists of one element, the zeroelement 0 of K. The same is true for Sol(S 0 ; K) if K does not contain elementsa with ap = 0 (for example, K is a field, or more general, K does not have zerodivisors). Thus the difference between X and Y becomes noticeable only if weadmit solutions with values in rings with zero divisors.

Corollary-Definition 1.4. Let X be an affine algebraic variety defined by a

system of algebraic equations S ⊂ k[T1 , . . . , Tn ]. The ideal (S) depends only onX and is called the defining ideal of X. It is denoted by I(X). For any idealI ⊂ k[T ] we denote by V (I) the affine algebraic k-variety corresponding to thesystem of algebraic equations I (or, equivalently, any set of generators of I).Clearly, the defining ideal of V (I) is I.

The next theorem is of fundamental importance. It shows that one can alwaysrestrict oneself to finite systems of algebraic equations.

Theorem 1.5. (Hilbert’s Basis Theorem). Let I be an ideal in the polynomial

Proof. The assertion is true if k[T ] is the polynomial ring in one variable. In fact,we know that in this case k[T ] is a principal ideal ring, i.e., each ideal is generatedby one element. Let us use induction on the number n of variables. Everypolynomial F (T ) ∈ I can be written in the form F (T ) = b0 Tnr + . . . + br , wherebi are polynomials in the first n−1 variables and b0 6= 0. We will say that r is thedegree of F (T ) with respect to Tn and b0 is its highest coefficient with respect toTn . Let Jr be the subset k[T1 , . . . , Tn−1 ] formed by 0 and the highest coefficientswith respect to Tn of all polynomials from I of degree r in Tn . It is immediatelychecked that Jr is an ideal in k[T1 , . . . , Tn−1 ]. By induction, Jr is generatedby finitely many elements a1,r , . . . , am(r),r ∈ k[T1 , . . . , Tn−1 ]. Let Fir (T ), i = 5

1, . . . , m(r), be the polynomials from I which have the highest coefficient equalto ai,r . Next, we consider the union J of the ideals Jr . By multiplying apolynomial F by a power of Tn we see that Jr ⊂ Jr+1 . This immediately impliesthat the union J is an ideal in k[T1 , . . . , Tn−1 ]. Let a1 , . . . , at be generatorsof this ideal (we use the induction again). We choose some polynomials Fi (T )which have the highest coefficient with respect to Tn equal to ai . Let d(i) bethe degree of Fi (T ) with respect to Tn . Put N = max{d(1), . . . , d(t)}. Let usshow that the polynomials

Fir , i = 1, . . . , m(r), r < N, Fi , i = 1, . . . , t,

where F 0 (T ) is of lower degree in Tn . Repeating this for F 0 (T ), if needed, we

obtain F (T ) ≡ R(T ) mod (F1 (T ), . . . , Ft (T )),where R(T ) is of degree d strictly less than N in Tn . For such R(T ) we cansubtract from it a linear combination of the polynomials Fi,d and decrease itsdegree in Tn . Repeating this, we see that R(T ) belongs to the ideal generatedby the polynomials Fi,r , where r < N . Thus F can be written as a linearcombination of these polynomials and the polynomials F1 , . . . , Ft . This provesthe assertion.

Finally, we define a subvariety of an affine algebraic variety.

Definition 1.2. An affine algebraic variety Y over k is said to be a subvarietyof an affine algebraic variety X over k if Y (K) ⊂ X(K) for any k-algebra K.We express this by writing Y ⊂ X. Clearly, every affine algebraic variety over k is a subvariety of some n-dimensionalaffine space Ank over k. The next result follows easily from the proof of Propo-sition 1.2:Proposition 1.6. An affine algebraic variety Y is a subvariety of an affine varietyX if and only if I(X) ⊂ I(Y ).6 LECTURE 1. SYSTEMS OF ALGEBRAIC EQUATIONS

define the same affine algebraic Q-varieties.

such that its set of K-solutions is equal to X(K) × X 0 (K) for any k-algebra K.We will denote it by X × Y and call it the Cartesian product of X and Y .4. Let X and X 0 be two subvarieties of Ank . Define an affine algebraic varietyover k such that its set of K-solutions is equal to X(K) ∩ X 0 (K) for any k-algebra K. It is called the intersection of X and X 0 and is denoted by X ∩ X 0 .Can you define in a similar way the union of two algebraic varieties?5. Suppose that S and S 0 are two systems of linear equations over a field k.Show that (S) = (S 0 ) if and only if Sol(S; k) = Sol(S 0 ; k).6. A commutative ring A is called Noetherian if every ideal in A is finitely gener-ated. Generalize Hilbert’s Basis Theorem by proving that the ring A[T1 , . . . , Tn ]of polynomials with coefficients in a Noetherian ring A is Noetherian.Lecture 2

Affine algebraic sets

Let X be an affine algebraic variety over k. For different k-algebras K the sets ofK-points X(K) could be quite different. For example it could be empty althoughX 6= ∅k . However if we choose K to be algebraically closed, X(K) is alwaysnon-empty unless X = ∅k . This follows from the celebrated Nullstellensatz ofHilbert that we will prove in this Lecture.Definition 2.1. Let K be an algebraically closed field containing the field k. Asubset V of K n is said to be an affine algebraic k-set if there exists an affinealgebraic variety X over k such that V = X(K). The field k is called the ground field or the field of definition of V . Sinceevery polynomial with coefficients in k can be considered as a polynomial withcoefficients in a field extension of k, we may consider an affine algebraic k-set asan affine algebraic K-set. This is often done when we do not want to specify towhich field the coefficients of the equations belong. In this case we call V simplyan affine algebraic set. First we will see when two different systems of equations define the sameaffine algebraic set. The answer is given in the next theorem. Before we stateit, let us recall that for every ideal I in a ring A its radical rad(I) is defined by rad(I) = {a ∈ A : an ∈ I for some n ≥ 0}.It is easy to verify that rad(I) is an ideal in A. Obviously, it contains I.Theorem 2.1. (Hilbert’s Nullstellensatz). Let K be an algebraically closed fieldand S and S 0 be two systems of algebraic equations in the same number ofvariables over a subfield k. Then Sol(S; K) = Sol(S 0 ; K) ⇐⇒ rad((S)) = rad((S 0 )).

78 LECTURE 2. AFFINE ALGEBRAIC SETS

Proof. Obviously, the set of zeroes of an ideal I and its radical rad(I) in K nare the same. Here we only use the fact that K has no zero divisors so thatF n (a) = 0 ⇐⇒ F (a) = 0. This proves ⇐. Let V be an algebraic set in K ngiven by a system of algebraic equations S. Let us show that the radical of theideal (S) can be defined in terms of V only:

G(a1 , . . . , an , an+1 ) = G(a1 , . . . , an ) = 0

will show that this implies that the ideal (Z) contains 1. Suppose this is true.Then, we may write X 1= PF F + Q(1 − Tn+1 G) F ∈S

for some polynomials PF and Q in T1 , . . . , Tn+1 . Plugging in 1/G instead of

Tn+1 and reducing to the common denominator, we obtain that a certain powerof G belongs to the ideal generated by the polynomials F, F ∈ S. So, we can concentrate on proving the following assertion:

Lemma 2.2. If I is a proper ideal in k[T ], then the set of its solutions in analgebraically closed field K is non-empty.

We use the following simple assertion which easily follows from the ZornLemma: every ideal in a ring is contained in a maximal ideal unless it coincideswith the whole ring. Let m be a maximal ideal containing our ideal I. We havea homomorphism of rings φ : k[T ]/I → A = k[T ]/m induced by the factor mapk[T ] → k[T ]/m . Since m is a maximal ideal, the ring A is a field containing 9

k as a subfield. Note that A is finitely generated as a k-algebra (because k[T ]

is). Suppose we show that A is an algebraic extension of k. Then we will beable to extend the inclusion k ⊂ K to a homomorphism A → K (since K isalgebraically closed), the composition k[T ]/I → A → K will give us a solutionof I in K n . Thus Lemma 2.2 and hence our theorem follows from the following:Lemma 2.3. Let A be a finitely generated algebra over a field k. Assume A isa field. Then A is an algebraic extension of k. Before proving this lemma, we have to remind one more definition fromcommutative algebra. Let A be a commutative ring without zero divisors (anintegral domain) and B be another ring which contains A. An element x ∈ B issaid to be integral over A if it satisfies a monic equation : xn +a1 xn−1 +. . .+an =0 with coefficients ai ∈ A. If A is a field this notion coincides with the notionof algebraicity of x over A. We will need the following property which will beproved later in Corollary 10.2. Fact: The subset of elements in B which are integral over A is a subring ofB. We will prove Lemma 2.3 by induction on the minimal number r of generatorst1 , . . . , tr of A. If r = 1, the map k[T1 ] → A defined by T1 7→ t1 is surjective. Itis not injective since otherwise A ∼ = k[T1 ] is not a field. Thus A ∼ = k[T1 ]/(F ) forsome F (T1 ) 6= 0, hence A is a finite extension of k of degree equal to the degreeof F . Therefore A is an algebraic extension of k. Now let r > 1 and supposethe assertion is not true for A. Then, one of the generators t1 , . . . , tr of A istranscendental over k. Let it be t1 . Then A contains the field F = k(t1 ), theminimal field containing t1 . It consists of all rational functions in t1 , i.e. ratios ofthe form P (t1 )/Q(t1 ) where P, Q ∈ k[T1 ]. Clearly A is generated over F by r−1generators t2 , . . . , tr . By induction, all ti , i 6= 1, are algebraic over F . We know d(i)that each ti , i 6= 1, satisfies an equation of the form ai ti +. . . = 0, ai 6= 0, wherethe coefficients belong to the field F . Reducing to the common denominator, wemay assume that the coefficients are polynomial in t1 , i.e., belong to the smallest d(i)−1subring k[t1 ] of A containing t1 . Multiplying each equation by ai , we seethat the elements ai ti are integral over k[t1 ]. At this point we can replace thegenerators ti by ai ti to assume that each ti is integral over k[t1 ]. Now using theFact we obtain that every polynomial expression in t2 , . . . , tr with coefficientsin k[t1 ] is integral over k[t1 ]. Since t1 , . . . , tr are generators of A over k, everyelement in A can be obtained as such polynomial expression. So every elementfrom A is integral over k[t1 ]. This is true also for every x ∈ k(t1 ). Since t110 LECTURE 2. AFFINE ALGEBRAIC SETS

is transcendental over k, k[x1 ] is isomorphic to the polynomial algebra k[T1 ].

Thus we obtain that every fraction P (T1 )/Q(T1 ), where we may assume thatP and Q are coprime, satisfies a monic equation X n + A1 X n + . . . + An = 0with coefficients from k[T1 ]. But this is obviously absurd. In fact if we plug inX = P/Q and clear the denominators we obtain

Next we shall show that the set of algebraic k-subsets in K n can be usedto define a unique topology in K n for which these sets are closed subsets. Thisfollows from the following:

Proposition 2.7. (i) The intersection ∩s∈S Vs of any family {Vs }s∈S of affine algebraic k-sets is an affine algebraic k-set in K n .

(ii) The union ∪s∈S Vs of any finite family of affine algebraic k-sets is an affine algebraic k-set in K n .

(iii) ∅ and K n are affine algebraic k-sets.

Proof. P (i) Let Is = I(Vs ) be the ideal of polynomials vanishing on Vs . LetI = s Is be the sum of the ideals Is , i.e., the minimal ideal of k[T ] containingthe sets Is . Since Is ⊂ I, we have V (I) ⊂ VP(Is ) = Vs . Thus V (I) ⊂ ∩s∈S Vs .Since each f ∈ I is equal to a finite sum fs , where fs ∈ Is , we see thatf vanishes at each x from the intersection. Thus x ∈ V (I), and we have theopposite inclusion. Q (ii) Let I be the ideal generated by products s fs , where fs ∈ Is . Ifx ∈ ∪s Vs , then x ∈ Vs for some s ∈ S. Hence all fs ∈ Is vanishes at x.But then all products vanishes at x, and therefore x ∈ V (I). This shows that∪s Vs ⊂ V (I). Conversely, suppose that all products vanish at x but x 6∈ Vs forany s. Then, for anyQ s ∈ S there exists some fs ∈ Is such that fs (x) 6= 0. Butthen the product s fs ∈ I does not vanish at x. This contradiction proves theopposite inclusion. (iii) This is obvious, ∅ is defined by the system {1 = 0}, K n is defined by thesystem {0 = 0}. Using the previous Proposition we can define the topology on K n by declaringthat its closed subsets are affine algebraic k- subsets. The previous propositionverifies the axioms. This topology on K n is called the Zariski k-topology (orZariski topology if k = K). The corresponding topological space K n is calledthe n-dimensional affine space over k and is denoted by Ank (K). If k = K, wedrop the subscript k and call it the n-dimensional affine space.

Example 2.8. A proper subset in A1 (K) is closed if and only if it is finite.

In fact, every ideal I in k[T ] is principal, so that its set of solutions coincideswith the set of solutions of one polynomial. The latter set is finite unless thepolynomial is identical zero.12 LECTURE 2. AFFINE ALGEBRAIC SETS

Remark 2.9. As the previous example easily shows the Zariski topology in K n isnot Hausdorff (=separated), however it satisfies a weaker property of separability.This is the property (T1 ): for any two points x 6= y in An (k), there exists an open subset U suchthat x ∈ U but y 6∈ U (see Problem 4). Any point x ∈ V = X(K) is defined by a homomorphism of k-algebrasevx : k[X]/I → K. Let p = Ker(evx ). Since K is a field, p is a prime ideal. Itcorresponds to a closed subset which is the closure of the set {x}. Thus, if x isclosed in the Zariski topology, the ideal p must be a maximal ideal. By Lemma2.3, in this case the quotient ring (k[X]/I)/px is an algebraic extension of k.Conversely, a finitely generated domain contained in an algebraic extension of kis a field (we shall prove it later in Lecture 10). Points x with the same idealKer(evx ) differ by a k-automorphism of K. Thus if we assume that K is analgebraically closed algebraic extension of k then all points of V are closed.

Morphisms of affine algebraic

varieties

In Lecture 1 we defined two systems of algebraic equations to be equivalent if

they have the same sets of solutions. This is very familiar from the theory oflinear equations. However this notion is too strong to work with. We can succeedin solving one system of equation if we would be able to find a bijective map ofits set of solutions to the set of solutions of another system of equations whichcan be solved explicitly. This idea is used for the following notion of a morphismbetween affine algebraic varieties.Definition 3.1. A morphism f : X → Y of affine algebraic varieties over a fieldk is a set of maps fK : X(K) → Y (K) where K runs over the set of k-algebrassuch that for every homomorphism of k-algebras φ : K → K 0 the followingdiagram is commutative: X(φ) X(K) / X(K 0 ) (3.1) FK fK 0 Y (K)

Y (K) / Y (K 0 )

We denote by MorAff /k (X, Y ) the set of morphisms from X to Y .

The previous definition is a special case of the notion of a morphism (or, anatural transformation) of functors. Let X be an affine algebraic variety. We know from Lecture 1 that for everyk-algebra K there is a natural bijection X(K) → Homk (k[T ]/I(X), K). (3.2)

1314 LECTURE 3. MORPHISMS OF AFFINE ALGEBRAIC VARIETIES

From now on we will denote the factor algebra k[T ]/I(X) by O(X) and willcall it the coordinate algebra of X. We can view the elements of this algebra asfunctions on the set of points of X. In fact, given a K-point a ∈ X(K) and anelement ϕ ∈ O(X) we find a polynomial P ∈ k[T ] representing ϕ and put

ϕ(a) = P (a).

Clearly this definition does not depend on the choice of the representative. An-other way to see this is to view the point a as a homomorphism eva : O(X) → K.Then ϕ(a) = eva (ϕ).Note that the range of the function ϕ depends on the argument: if a is a K-point,then ϕ(a) ∈ K. Let ψ : A → B be a homomorphism of k-algebras. For every k-algebra Kwe have a natural map of sets Homk (B, K) → Homk (A, K), which is obtainedby composing a map B → K with ψ. Using the bijection (3.2) we see that anyhomomorphism of k-algebras

ψ : O(Y ) → O(X)

defines a morphism f : X → Y by setting, for any α : O(X) → K,

fK (α) = α ◦ ψ. (3.3)

Thus we have a natural map of sets

ξ : Homk (O(Y ), O(X)) → MorAff /k (X, Y ). (3.4)

Recall how this correspondence works. Take a K-point a = (a1 , . . . , an ) ∈

X(K) in a k-algebra K. It defines a homomorphism

eva : O(X) = k[T1 , . . . , Tn ]/I(X) → K

by assigning ai to Ti , i = 1, . . . , n. Composing this homomorphism with a

Take the identity map idO(X) in the left bottom set. It goes to the element αin the left top set. The bottom horizontal arrow sends idO(X) to ψ. The rightvertical arrow sends it to α ◦ ψ. Now, because of the commutativity of thediagram, this must coincide with the image of α under the top arrow, which isfK (α). This proves the surjectivity. The injectivity is obvious.

As soon as we know what is a morphism of affine algebraic k-varieties we

know how to define an isomorphism. This will be an invertible morphism. Weleave to the reader to define the composition of morphisms and the identitymorphism to be able to say what is the inverse of a morphism. The followingproposition is clear.

Proposition 3.2. Two affine algebraic k-varieties X and Y are isomorphic if

and only if their coordinate k-algebras O(X) and O(Y ) are isomorphic.

Note that it suffices to check the previous condition only for generators of theideal I(Y ), for example for the polynomials defining the system of equations Y .In terms of the polynomials (P1 (T ), . . . , Pm (T )) satisfying (3.5), the morphismf : X → Y is given as follows:

fK (a) = (P1 (a), . . . , Pm (a)) ∈ Y (K), ∀a ∈ X(K).

It follows from the definitions that a morphism φ given by polynomials

its value at T1 , i.e. by an element of O(X). This gives us one more interpretationof the elements of the coordinate algebra O(X). This time they are morphismsfrom X to A1k and hence again can be thought as functions on X. Let f : X → Y be a morphism of affine algebraic varieties. We know that itarises from a homomorphism of k-algebras f ∗ : O(Y ) → O(X).

Proposition 3.4. For any ϕ ∈ O(Y ) = MorAff /k (Y, A1k ),

f ∗ (ϕ) = ϕ ◦ f.

Proof. This follows immediately from the above definitions.

This justifies the notation f ∗ (the pull-back of a function).

By now you must feel comfortable with identifying the set X(K) of K-solutions of an affine algebraic k-variety X with homomorphisms O(X) →K. The identification of this set with a subset of K n is achieved by choos-ing a set of generators of the k-algebra O(X). Forgetting about generatorsgives a coordinate-free definition of the set X(K). The correspondence K →Hom( O(X), K) has the property of naturality, i.e. a homomorphism of k-algebras K → K 0 defines a map Homk (O(X), K) → Homk (O(X), K 0 ) suchthat a natural diagram, which we wrote earlier, is commutative. This leads to ageneralization of the notion of an affine k-variety.

Definition 3.2. An (abstract) affine algebraic k-variety is the correspondence

which assigns to each k-algebra K a set X(K). This assignment must satisfythe following properties:

(i) for each homomorphism of k-algebras φ : K → K 0 there is a map X(φ) :

(ii) X(idK ) = idX(K) ;

(iv) there exists a finitely generated k-algebra A such that for each K there is a bijection X(K) → Homk (A, K) and the maps X(φ) correspond to the composition maps Hom( A, K) → Homk (A, K 0 ).18 LECTURE 3. MORPHISMS OF AFFINE ALGEBRAIC VARIETIES

We leave to the reader to define a morphism of abstract affine algebraic k-

varieties and prove that they are defined by a homomorphism of the correspondingalgebras defined by property (iii). A choice of n generators f1 , . . . , fn ) of Adefines a bijection from X(K) to a subset Sol(I; K) ⊂ K n , where I is thekernel of the homomorphism k[T1 , . . . , Tn ] → A, defined by Ti 7→ fi . Thisbijection is natural in the sense of the commutativity of the natural diagrams.

algebraic k-variety. The corresponding algebra A is k[T ]/(S).5. The correspondence K → K ∗ ( = invertible elements in K) is an abstractaffine algebraic k-variety. The corresponding algebra A is equal to k[T1 , T2 ]/(T1 T2 −1). The cosets of T1 and T2 define a set of generators such that the correspondingaffine algebraic k-variety is a subvariety of A2 . It is denoted by Gm,k and is calledthe multiplicative algebraic group over k. Note that the maps X(K) → X(K 0 )are homomorphisms of groups.6. More generally we may consider the correspondence K → GL(n, K) (=invert-ible n × n matrices with entries in K). It is an abstract affine k-variety definedby the quotient algebra k[T11 , . . . , Tnn , U ]/(det((Tij )U − 1). It is denoted byGLk (n) and is called the general linear group of degree n over k.

Remark 3.6. We may make one step further and get rid of the assumption in(iv) that A is a finitely generated k-algebra. The corresponding generalization iscalled an affine k-scheme. Note that, if k is algebraically closed, the algebraic setX(k) defined by an affine algebraic k-variety X is in a natural bijection with theset of maximal ideals in O(X). This follows from Corollary 2.6 of the Hilbert’sNullstellensatz. Thus the analog of the set X(k) for the affine scheme is theset Spm(A) of maximal ideals in A. For example take an affine scheme definedby the ring of integers Z. Each maximal ideal is a principal ideal generated bya prime number p. Thus the set X(k) becomes the set of prime numbers. Anumber m ∈ Z becomes a function on the set X(k). It assigns to a primenumber p the image of m in Z/(p) = Fp , i.e., the residue of m modulo p. Now, we specialize the notion of a morphism of affine algebraic varieties todefine the notion of a regular map of affine algebraic sets. Recall that an affine algebraic k-set is a subset V of K n of the form X(K),where X is an affine algebraic variety over k and K is an algebraically closedextension of k. We can always choose V to be equal V (I),where I is a radicalideal. This ideal is determined uniquely by V and is equal to the ideal I(V )of polynomials vanishing on V (with coefficients in k). Each morphism f : 19

Irreducible algebraic sets and

rational functions

We know that two affine algebraic k-sets V and V 0 are isomorphic if and only iftheir coordinate algebras O(V ) and O(V 0 ) are isomorphic. Assume that both ofthese algebras are integral domains (i.e. do not contain zero divisors). Then theirfields of fractions R(V ) and R(V 0 ) are defined. We obtain a weaker equivalenceof varieties if we require that the fields R(V ) and R(V 0 ) are isomorphic. Inthis lecture we will give a geometric interpretation of this equivalence relation bymeans of the notion of a rational function on an affine algebraic set. First let us explain the condition that O(V ) is an integral domain. We recallthat V ⊂ K n is a topological space with respect to the induced Zariski k-topology of K n . Its closed subsets are affine algebraic k-subsets of V . Fromnow on we denote by V (I) the affine algebraic k-subset of K n defined by theideal I ⊂ k[T ]. If I = (F ) is the principal ideal generated by a polynomial F , wewrite V ((F )) = V (F ). An algebraic subset of this form, where (F ) 6= (0), (1),is called a hypersurface.Definition 4.1. A topological space V is said to be reducible if it is a union oftwo proper non-empty closed subsets (equivalently, there are two open disjointproper subsets of V ). Otherwise V is said to be irreducible. By definition theempty set is irreducible. An affine algebraic k-set V is said to be reducible (resp.irreducible) if the corresponding topological space is reducible (resp. irreducible).Remark 4.1. Note that a Hausdorff topological space is always reducible unless itconsists of at most one point. Thus the notion of irreducibility is relevant only fornon-Hausdorff spaces. Also one should compare it with the notion of a connected

2122LECTURE 4. IRREDUCIBLE ALGEBRAIC SETS AND RATIONAL FUNCTIONS

space. A topological spaces X is connected if it is not equal to the union of two

disjoint proper closed (equivalently open) subsets. Thus an irreducible space isalways connected but the converse is not true in general. For every affine algebraic set V we denote by I(V ) the ideal of polynomialsvanishing on V . Recall that, by Nullstellensatz, I(V (I)) = rad(I).

Proposition 4.2. An affine algebraic set V is irreducible if and only if its coor-dinate algebra O(V) has no zero divisors.

Theorem 4.4. 1. Let V be a Noetherian topological space. Then V is a union

Proof. Let us prove the first part. If V is irreducible, then the assertion is obvious.Otherwise, V = V1 ∪ V2 , where Vi are proper closed subsets of V . If both ofthem are irreducible, the assertion is true. Otherwise, one of them, say V1 isreducible. Hence V1 = V11 ∪ V12 as above. Continuing in this way, we either stopsomewhere and get the assertion or obtain an infinite strictly decreasing sequenceof closed subsets of V . The latter is impossible because V is Noetherian. Toprove the second assertion, we assume that

Since V1 is irreducible, one of the subsets V1 ∩ Wj is equal to V1 , i.e., V1 ⊂ Wj .

We may assume that j = 1. Similarly, we show that W1 ⊂ Vi for some i. HenceV1 ⊂ W1 ⊂ Vi . This contradicts the assumption Vi 6⊂ Vj for i 6= j unlessV1 = W1 . Now we replace V by V2 ∪ . . . ∪ Vk = W2 ∪ . . . ∪ Wt and repeat theargument. An irreducible closed subset Z of a topological space X is called an irreduciblecomponent if it is not properly contained in any irreducible closed subset. Let Vbe a Noetherian topological space and V = ∪i Vi , where Vi are irreducible closedsubsets of V with Vi 6⊂ Vj for i 6= j, then each Vi is an irreducible component.Otherwise Vi is contained properly in some Z, and Z = ∪i (Z ∩ Vi ) would implythat Z ⊂ Vi for some i hence Vi ⊂ Vk . The same argument shows that everyirreducible component of X coincides with one of the Vi ’s.Remark 4.5. Compare this proof with the proof of the theorem on factorizationof integers into prime factors. Irreducible components play the role of primefactors. In view of Proposition 4.3, we can apply the previous terminology to affinealgebraic sets V . Thus, we can speak about irreducible affine algebraic k-sets,irreducible components of V and a decomposition of V into its irreducible compo-nents. Notice that our topology depends very much on the field k. For example,an irreducible k-subset of K is the set of zeroes of an irreducible polynomial ink[T ]. So a point a ∈ K is closed only if a ∈ k. We say that V is geometricallyirreducible if it is irreducible considered as a K-algebraic set. Recall that a polynomial F (T ) ∈ k[T ] is said to be irreducible if F (T ) =G(T )P (T ) implies that one of the factors is a constant (since k[T ]∗ = k ∗ , this24LECTURE 4. IRREDUCIBLE ALGEBRAIC SETS AND RATIONAL FUNCTIONS

is equivalent to saying that F (T ) is an irreducible or prime element of the ring

k[T ]).Lemma 4.6. Every polynomial F ∈ k[T1 , . . . , Tn ] is a product of irreduciblepolynomials which are defined uniquely up to multiplication by a constant.Proof. This follows from the well-known fact that the ring of polynomials k[T1 , . . . , Tn ]is a UFD (a unique factorization domain). The proof can be found in any ad-vanced text-book of algebra.Proposition 4.7. Let F ∈ k[T ]. A subset Z ⊂ K n is an irreducible componentof the affine algebraic set V = V (F ) if and only if Z = V (G) where G isan irreducible factor of F . In particular, V is irreducible if and only if F is anirreducible polynomial.Proof. Let F = F1a1 . . . Frar be a decomposition of F into a product of irreduciblepolynomials. Then V (F ) = V (F1 ) ∪ . . . ∪ V (Fr )and it suffices to show that V (Fi ) is irreducible for every i = 1, . . . , r. Moregenerally, we will show that V (F ) is irreducible if F is irreducible. By Proposition4.2, this follows from the fact that the ideal (F ) is prime. If (F ) is not prime,then there exist P, G ∈ k[T ] \ (F ) such that P G ∈ (F ). The latter implies thatF |P G. Since F is irreducible, F |P or F |G (this follows easily from Lemma 4.6).This contradiction proves the assertion. Let V ⊂ K n be an irreducible affine algebraic k-set and O(V ) be its coordi-nate algebra. By Proposition 4.2, O(V ) is a domain, therefore its quotient fieldQ(O(V )) is defined. We will denote it by R(V ) and call it the field of rationalfunctions on V . Its elements are called rational functions on V . Recall that for every integral domain A its quotient field Q(A) is a fielduniquely determined (up to isomorphisms) by the following two conditions: (i) there is an injective homomorphism of rings i : A → Q(A); (ii) for every injective homomorphism of rings φ : A → K, where K is a field, there exists a unique homomorphism φ̄ : Q(A) → K such that φ̄ ◦ i = φ.The field Q(A) is constructed as the factor-set A × (A \ {0})/R , where R isthe equivalence relation (a, b) ∼ (a0 , b0 ) ⇐⇒ ab0 = a0 b. Its elements are denotedby ab and added and multiplied by the rules a a0 ab0 + a0 b a a0 aa0 + 0 = , · 0 = 0. b b bb0 b b bb 25

topology, P ≡ 0 on V , i.e. P ∈ I(V ). This shows that under the map R(W ) →R(V ), F goes to 0. Since the homomorphism R(W ) → R(V ) is injective (anyhomomorphism of fields is injective) this is absurd. In particular, taking W = A1k (K), we obtain the interpretation of elementsof the field R(V ) as non-constant rational functions V − → K defined on anopen subset of V (the complement of the set of the zeroes of the denominator).From this point of view, the homomorphism R(W ) → R(V ) defining a rationalmap f : V − → W can be interpreted as the homomorphism f ∗ defined by thecomposition φ 7→ φ ◦ f .

that char(k) 6= 2. A rational map f : V − → W is given by a homomorphismf ∗ : R(W ) → R(V ). Restricting it to O(W ) and composing it with k[T1 , T2 ] →O(W ), we obtain two rational functions R1 (T ) and R2 (T ) such that R1 (T )2 +R2 (T )2 = 1 (they are the images of the unknowns T1 and T2 ). In other words,we want to find “a rational parameterization” of the circle, that is, we wantto express the coordinates (t1 , t2 ) of a point lying on the circle as a rationalfunction of one parameter. It is easy to do this by passing a line through thispoint and the fixed point on the circle, say (1, 0). The slope of this line is theparameter associated to the point. Explicitly, we write T2 = T (T1 − 1), plug intothe equation T12 + T22 = 1 and find

T2 − 1 −2T T1 = 2 , T2 = 2 . T +1 T +1 27

Thus, our rational map is given by

R(V (T12 + T22 − 1)) ∼

= k(T1 ).

The next theorem, although sounding as a deep result, is rather useless forconcrete applications.Theorem 4.10. Assume k is of characteristic 0. Then any irreducible affinealgebraic k-set is birationally isomorphic to an irreducible hypersurface.Proof. Since R(V ) is a finitely generated field over k, it can be obtained as analgebraic extension of a purely transcendental extension L = k(t1 , . . . , tn ) of k.Since char(k) = 0, R(V ) is a separable extension of L, and the theorem on aprimitive element applies (M. Artin, ”Algebra”, Chapter 14, Theorem 4.1): analgebraic extension K/L of characteristic zero is generated by one element x ∈K. Let k[T1 , . . . , Tn+1 ] → R(V ) be defined by sending Ti to ti for i = 1, . . . , n,and Tn+1 to x. Let I be the kernel, and φ : A = k[T1 , . . . , Tn+1 ]/I → R(V )be the corresponding injective homomorphism. Every P (T1 , . . . , Tn+1 ) ∈ I ismapped to P (t1 , . . . , tn , x) = 0. Considering P (x1 , . . . , xn , Tn+1 ) as an elementof L[Tn+1 ] it must be divisible by the minimal polynomial of x. Hence I =(F (T1 , . . . , Tn , Tn+1 )), where F (t1 , . . . , tn , Tn+1 ) is a product of the minimalpolynomial of x and some polynomial in t1 , . . . , tn . Since A is isomorphic toa subring of a field it must be a domain. By definition of the quotient field φcan be extended to a homomorphism of fields Q(A) → R(V ). Since R(V ) isgenerated as a field by elements in the image, φ must be an isomorphism. ThusR(V ) is isomorphic to Q(k[T1 , . . . , Tn+1 ]/(F )) and we are done.Remark 4.11. The assumption char(k) = 0 can be replaced by the weaker as-sumption that k is a perfect field, for example, k is algebraically closed. In thiscase one can show that R(V ) is a separable extension of some purely transcen-dental extension of k.28LECTURE 4. IRREDUCIBLE ALGEBRAIC SETS AND RATIONAL FUNCTIONS

Definition 4.5. An irreducible affine algebraic k-set V is said to be k-rational

if R(V ) ∼ = k(T1 , . . . , Tn ) for some n. V is called rational if, viewed as algebraicK-set, it is K-rational.Example 4.12. 2. Assume char(k) 6= 2. The previous example shows that thecircle V (T12 + T22 − 1) is k-rational for any k. On the other hand, V (T12 + T22 + 1)may not be k-rational, for example, when k = R.3. An affine algebraic set given by a system of linear equations is always rational(Prove it!).4. V (T12 + T23 − 1) is not rational. Unfortunately, we do not have yet sufficienttools to show this.5. Let V = V (T13 + . . . + Tn3 − 1) be a cubic hypersurface. It is known that Vis not rational for n = 2 and open question for many years whether V is rationalfor n = 4. The negative answer to this problem was given by Herb Clemens andPhillip Griffiths in 1972. It is known that V is rational for n ≥ 5 however it isnot known whether V (F ) is rational for any irreducible polynomial of degree 3in n ≥ 5 variables. An irreducible algebraic set V is said to be k-unirational if its field of rationalfunctions R(V ) is isomorphic to a subfield of k(T1 , . . . , Tn ) for some n. Itwas an old problem (the Lüroth Problem) whether, for k = C, there exist k-unirational sets which are not k-rational. The theory of algebraic curves easilyimplies that this is impossible if C(V ) is transcendence degree 1 over C. A purelyalgebraic proof of this fact is not easy (see P. Cohn, “Algebra”). The theory ofalgebraic surfaces developed in the end of the last century by Italian geometersimplies that this is impossible if C(V ) of transcendence degree 2 over C. Nopurely algebraic proofs of this fact is known. Only in 1972-73 a first exampleof a unirational non-rational set was constructed. In fact, there were givenindependently 3 counterexamples (by Clemens-Griffiths, by Artin-Mumford andIskovskih-Manin). The example of Clemens-Griffiths is the cubic hypersurfaceV (T13 + T23 + T33 + T43 − 1). Finally we note that we can extend all the previous definitions to the case ofaffine algebraic varieties. For example, we say that an affine algebraic variety Xis irreducible if its coordinate algebra O(X) is an integral domain. We leave tothe reader to do all these generalizations.

Projective algebraic varieties

Let A be a commutative ring and An+1 (n ≥ 0) be the Cartesian product

equipped with the natural structure of a free A-module of rank n + 1. A freesubmodule M of An+1 of rank 1 is said to be a line in An+1 , if M = Ax forsome x = (a0 , . . . , an ) such that the ideal generated by a0 , . . . , an contains 1.We denote the set of lines in An+1 by Pn (A)0 . One can define Pn (A)0 also asfollows. Let

C(A)n = {x = (a0 , . . . , an ) ∈ An+1 : (a0 , . . . , an ) = 1}.

Then each line is generated by an element of C(A)n . Two elements x, y ∈ C(A)n

define the same line if and only if x = λy for some invertible λ ∈ A. Thus

Pn (A)0 = C(A)n /A∗ ,

is the set of orbits of the group A∗ of invertible elements of A acting on C(A)n

C(A)n = An+1 \ {0}, Pn (A)0 = (An+1 \ {0})/A∗ .

homogeneous coordinates of the line. In view of the above they are determineduniquely up to an invertible scalar factor λ ∈ A∗ .Example 5.1. 1. Take A = R. Then P1 (R)0 is the set of lines in R2 passingthrough the origin. By taking the intersection of the line with the unit circlewe establish a bijective correspondence between P1 (R) and the set of pointson the unit circle with the identification of the opposite points. Or choosing a

3132 LECTURE 5. PROJECTIVE ALGEBRAIC VARIETIES

representative on the upper half circle we obtain a bijective map from P1 (R)0 tothe half circle with the two ends identified. This is bijective to a circle. Similarlywe can identify P2 (R)0 with the set of points in the upper unit hemisphere suchthat the opposite points on the equator are identified. This is homeomorphicto the unit disk where the opposite points on the boundary are identified. Theobtained topological space is called the real projective plane and is denoted byRP2 .2. Take A = C. Then P1 (C)0 is the set of one-dimensional linear subspacesof C2 . We can choose a unique basis of x ∈ P1 (C)0 of the form (1, z) unlessx = (0, z), z ∈ C \ {0}, and Cx = C(0, 1). In this way we obtain a bijective mapfrom P1 (C)0 to c ∪ {∞}, the extended complex plane. Using the stereographicprojection, we can identify the latter set with a 2-dimensional sphere. The com-plex coordinates make it into a compact complex manifold of dimension 1, theRiemann sphere CP1 .

So each line is a direct summand of An+1 . Not each direct summand of An+1 isnecessarily free. So we can enlarge the set Pn (A)0 by adding to it not necessarilyfree direct summands of An+1 which become free of rank 1 after “localizing” thering. Let us explain the latter. Let S be a non-empty multiplicatively closed subset of A containing 1. Onedefines the localization MS of an A-module M in the similar way as one defines 33

the field of fractions: it is the set of equivalence classes of pairs (m, s) ∈ M ×

S with the equivalence relation: (m, s) ≡ (m0 , s0 ) ⇐⇒ ∃s00 ∈ S such thats00 (s0 m − sm0 ) = 0. The equivalence class of a pair (m, s) is denoted by ms . Theequivalence classes can be added by the natural rule m m0 s0 m + sm0 + 0 = s s ss0(one verifies that this definition is independent of a choice of a representative).If M = A, one can also multiply the fractions by the rule a a0 aa0 · 0 = . s s ssThus AS becomes a ring such that the natural map A → AS , a 7→ a1, is ahomomorphism of rings. The rule a m am · 0 = 0. s s ssequips MS with the structure of an AS -module. Note that MS = {0} if 0 ∈ S.Observe also that there is a natural isomorphism of AS -modules a am M ⊗A AS → MS , m ⊗ 7→ , s swhere AS is equipped with the structure of an A-module by means of the canon-ical homomorphism A → AS .Example 5.2. 3. Take S to be the set of elements of A which are not zero-divisors. This is obviously a multiplicatively closed subset of A. The localizedring AS is called the total ring of fractions. If A is a domain, S = A \ {0}, andwe get the field of fractions. 4. Let p be a prime ideal in A. By definition of a prime ideal, the set A \ p ismultiplicatively closed. The localized ring AA\p is denoted by Ap and is called thelocalization of A at a prime ideal p. For example, take A = Z and p = (p), where pis a prime number. The ring Z(p) is isomorphic to the subring of Q which consists offractions such that the denominator is not divisible by p.

As we saw earlier any line L = Ax ∈ Pn (A)0 is a direct summand of the free

module An+1 . In general not every direct summand of a free module is free.Definition 5.1. A projective module over A is a finitely generated module Pover A satisfying one of the following equivalent properties:34 LECTURE 5. PROJECTIVE ALGEBRAIC VARIETIES

(i) P is isomorphic to a direct summand of a free module;

(ii) For every surjective homomorphism φ : M → P of A-modules there is a

homomorphism s : P → M such that φ ◦ s = idP (a section).

Let us prove the equivalence.

(ii)⇒ (i) Let An → P be the surjective homomorphism corresponding to achoice of generators of P . By property(i) there is a homomorphism s : P → Ansuch that φ ◦ s = idP . Let N = Ker(φ). Consider the homomorphism (i, s) :N ⊕ P → An , where i is the identity map N → An . It has the inverse given bym 7→ (m − φ(m), φ(m)) (i)⇒ (ii) Assume P ⊕N ∼ = An . Without loss of generality we may assume that nP, N are submodules of A . Let φ : M → P be a surjective homomorphism ofA-modules. We extend it to a surjective homomorphism (φ, idN ) : M ⊕N → An .If we prove property (ii) for free modules, we will be done since the restrictionof the corresponding section to P is a section of φ. So let φ : M → An be asurjective homomorphism. Let m1 , . . . , mn be some pre-images of the elementsof a basis (ξ1 , . . . , ξn ) of An . The homomorphism An → M defined by ξ 7→ miis well-defined and is a section. We saw in the previous proof that a free finitely generated module is projec-tive. In general, the converse is not true. For example, let K/Q be a finite fieldextension, and A be the ring of integers of K, i.e. the subring of elements of Kwhich satisfy a monic equation with coefficients in Z. Then any ideal in A is aprojective module but not necessarily a principal ideal. An important class of rings A such that any projective module over A is freeis the class of local rings. A commutative ring is called local if it has a unique maximal ideal. Forexample, any field is local. The ring of power series k[[T1 , . . . , Tn ]] is local (themaximal ideal is the set of infinite formal series with zero constant term).

Lemma 5.3. Let A be a local ring and m be its unique maximal ideal. ThenA \ m = A∗ (the set of invertible elements in A).

Proof. Let x ∈ A \ m. Then the principal ideal (x) is contained in some propermaximal ideal unless (x) = A which is equivalent to x ∈ A∗ . Since A has onlyone maximal ideal and it does not contain x, we see that (x) = A.

Proposition 5.4. A projective module over a local ring is free.

35

Proof. Let Matn (A) be the ring of n × n matrices with coefficients in a commu-tative ring A. For any ideal I in A we have a natural surjective homomorphismof rings Matn (A) → Matn (A/I), X 7→ X̄, which obtained by replacing eachentry of a matrix X with its residue modulo I. Now let A be a local ring, I = mbe its unique maximal ideal, and k = A/m (the residue field of A). SupposeX ∈ Matn (A) is such that X̄ is an invertible matrix in Matn (k). I claim thatX is invertible in Matn (A). In fact, let Ȳ · X̄ = In for some Y ∈ Matn (A).The matrix Y X has diagonal elements congruent to 1 modulo m and all off-diagonal elements belonging to m. By Lemma 5.3, the diagonal elements ofY X are invertible in A. It is easy to see, that using elementary row trans-formations which do not involve switching the rows we can reduce Y X to theidentity matrix. This shows that there exists a matrix S ∈ Matn (A) such thatS(Y X) = (SY )X = In . Similarly, using elementary column transformations,we show that X has the right inverse, and hence is invertible. Let M be a A-module and I ⊂ A an ideal. Let IM denote the submoduleof M generated by all products am, where a ∈ I. The quotient module M =M/IM is a A/I-module via the scalar multiplication (a + I)(m + IM ) = am +IM . There is an isomorphism of A/I-modules M/IM ∼ = M ⊗ M ⊗A (A/I),where A/I is considered as an A-algebra via the natural homomorphism A →A/I. It is easy to check the following property. (M ⊕ N )/I(M ⊕ N ) ∼ = (M/IM ) ⊕ (N/IN ). (5.1)Now let M be a projective module over a local ring A. Replacing M by anisomorphic module we may assume that M ⊕ N = An for some submodule Nof a free A-module An . Let m be the maximal ideal of A. Let (m1 , . . . , ms )be elements in M such that (m1 + I, . . . , ms + I) is a basis of the vector spaceM/mM over k = A/m. Similarly, choose (n1 , . . . , nt ) in N . By property (5.1)the residues of m1 , . . . , mt , n1 , . . . , ns form a basis of k n . Consider the mapf : An → M ⊕ N defined by sending the unit vector ei ∈ An to mi if i ≤ t andto ni if i ≥ t + 1. Let S be its matrix with respect to the unit bases (e1 , . . . , en )in An . Then the image of S in Matn (k) is an invertible matrix. Therefore S isan invertible matrix. Thus f is an isomorphism of A-modules. The restriction off to the free submodule Ae1 + . . . + Aet is an isomorphism At ∼ = M.Corollary 5.5. Let P be a projective module over a commutative ring A. Forany maximal ideal m in A the localization Pm is a free module over Am .Proof. This follows from the following lemma which we leave to the reader toprove.36 LECTURE 5. PROJECTIVE ALGEBRAIC VARIETIES

Lemma 5.6. Let P be a projective module over A. For any A-algebra B thetensor product P ⊗A B is a projective B-module.Definition 5.2. A projective module P over A has rank r if for each maximalideal m the module Pm is free of rank r.Remark 5.7. Note that, in general, a projective module has no rank. For example,let A = A1 × A2 be the direct sum of rings. The module Ak1 × An2 (with scalarmultiplication (a1 , a2 ) · (m1 , m2 ) = (a1 m1 , a2 m2 )) is projective but has no rankif k 6= n. If A is a domain, then the homomorphism A → Am defines anisomorphism of the fields of fractions Q(A) ∼ = Q(Am ). This easily implies thatthe rank of P can be defined as the dimension of the vector space P ⊗A Q(A). We state without proof the converse of the previous Corollary (see, for ex-ample, N. Bourbaki, “Commutative Algebra”, Chapter 2, §5).Proposition 5.8. Let M be a module over A such that for each maximal idealm the module Mm is free. Then M is a projective module. Now we are ready to give the definition of Pn (A).Definition 5.3. Let A be any commutative ring. The projective n-space over Ais the set Pn (A) of projective A-modules of rank 1 which are direct summandsof An+1 . We have seen that Pn (A)0 ⊂ Pn (A).The difference is the set of non-free projective modules of rank 1 which are directsummands of An+1 . A projective submodule of rank 1 of An+1 may not be a direct summand. Forexample, a proper principal ideal (x) ⊂ A is not a direct summand in A. A freesubmodule M = A(a0 , . . . , an ) of An+1 of rank 1 is a direct summand if andonly if the ideal generated by a0 , . . . , an is equal to A, i.e. M ∈ Pn (A)0 . This follows from the following characterization of direct summands of An+1 .A submodule M of An+1 is a direct summand if and only if the correspondinghomomorphism of the dual modules

An+1 ∼ = HomA (An+1 , A) → M ∗ = HomA (M, A)

is surjective. Sometimes Pn (A) is defined in “dual terms” as the set of projective

modules of rank 1 together with a surjective homomorphism An+1 → M . When 37

A is a field this is a familiar duality between lines in a vector space V and

hyperplanes in the dual vector space V ∗ . A set {fi }i∈I of elements from A is called a covering set if it generatesthe unit P ideal. Every covering set contains a finite covering subset. In fact if1 = i ai fi for some ai ∈ A, we choose those fi which occur in this sum withnon-zero coefficient. For any f ∈ A we set Af = AS , where S consists of powersof f .Lemma 5.9. Let M be a projective module of rank r over a ring A. Thereexists a finite covering set {fi }i∈I of elements in A such that for any i ∈ I thelocalization Mfi is a free Afi -module of rank r.Proof. We know that for any maximal ideal m in A the localization Mm is afree module of rank r. Let x1 , . . . , xr be its generators. Multiplying them byinvertible elements in Am , we may assume that the generators belong to A. Letφ : Ar → M be the homomorphism defined by these generators. We know thatthe corresponding homomorphism φm : Arm → Mm of localizations is bijective.I claim that there exists fm 6∈ m such the homomorphism φfm : Arfm → Mfm isbijective. Let K = Ker(φ) and C = Coker(φ). Then Ker(φm ) = Km = {0}.Since K is a finitely generated module and Km = 0, there exists g 6∈ m suchgK = {0} and hence Kg = 0. Similarly, we find an element h 6∈ m suchthat Ch = {0}. Now if we take fm = gh, then Kf and Cfm = {0}, henceφfm : Arfm → Mfm is bijective. Since the set of elements fm is not contained inany maximal ideal, it must generate the unit ideal, hence it is a covering set. Itremains to select a finite covering subset of the set {fm } .

Using Lemma 5.9 we may view every projective submodule M of An+1

of rank 1 as a ‘local line’ : we can find a finite covering set {fi }i∈I suchthat Mfi is a line in (Afi )n+1 . We call such a family a trivializing family forM . If {gj }j∈J is another trivializing family for M we may consider the family{fi gjP }(i,j)∈I×J . It P is a covering family as one sees by multiplying the two relations1 = i ai fi , 1 = j bj gj . Note that for any f, g ∈ A there is a natural homo-morphism of rings Af → Af g , a/f n → ag n /(f g)n inducing an isomorphism ofAf g -modules Mf ⊗Af Af g ∼ = Mf g . This shows that {fi gj }(i,j)∈I×J is a trivializingfamily. Moreover, if Mfi = xi Afi , xi ∈ An+1 fi and Mgj = yj Agj , yj ∈ An+1 gj , then

x0i = αij yj0 for some αij ∈ Afi gj (5.2)

Now let us go back to algebraic equations. Fix a field k. For any k-algebraK we have the set Pn (K). It can be viewed as a natural extension (in n + 1different ways) of the set Ank (K) = K n . In fact, for every k-algebra K we havethe injective maps αi : Ank (K) = K n → Pnk (K), (a1 , . . . , an ) → (a1 , . . . , ai , 1, ai+1 , . . . , an ), i = 0, . . . , n. Assume that K is a field. Take, for example, i = 0. We see that

Pn (K) \ K n = {(a0 , a1 , . . . , an )A ∈ Pn (K) : a0 = 0}.

It is naturally bijectively equivalent to Pn−1 (K). Thus we have

a Pn (K) = Ank (K) Pn−1 (K).

By now, I am sure you understand what I mean when I say “naturally”. Thebijections we establish for different K are compatible with respect to the mapsPn (K) → Pn (K 0 ) and K n → K 0n corresponding to homomorphisms K → K 0of k-algebras.

Example 5.10. The Riemann sphere

P1 (C) = C ∪ {P0 (C)}.

The real projective plane

P2 (R) = R2 ∪ P1 (R).

We want to extend the notion of an affine algebraic variety by considering

solutions of algebraic equations which are taken from Pn (K). Assume first thatL ∈ Pn (K) is a global line, i.e. a free submodule of K n+1 . Let (a0 , . . . , an ) be itsgenerator. For any F ∈ k[T0 , . . . , Tn ] it makes sense to say that F (a0 , . . . , an ) =0. However, it does not make sense, in general, to say that F (L) = 0 becausea different choice of a generator may give F (a0 , . . . , an ) 6= 0. However, we cansolve this problem by restricting ourselves only with polynomials satisfying

F (λT0 , . . . , λTn ) = λd F (T0 , . . . , Tn ), ∀λ ∈ K ∗ .

To have this property for all possible K, we require that F be a homogeneous

with |i| = d for all i. Here we use the vector notation for polynomials: i = (i0 , . . . , in ) ∈ Nn+1 , Ti = T0i0 · · · Tnin , |i| = i0 + . . . + in .By definition the constant polynomial 0 is homogeneous of any degree. Equivalently, F is homogeneous of degree d if the following identity in thering k[T0 , . . . , Tn , t] holds: F (tT0 , . . . , tTn ) = td F (T0 , . . . , Tn ). Let k[T ]d denote the set of all homogeneous polynomials of degree d. Thisis a vector subspace over k in k[T ] and k[T ] = ⊕d≥0 k[T ]d .Indeed every polynomial can be written uniquely as a linear combination of mono-mials Ti which are homogeneous of degree |i|. We write degF = d if F is ofdegree d. Let F be homogeneous polynomial in T0 , . . . , Tn . For any k-algebra K andx ∈ K n+1 F (x) = 0 ⇐⇒ F (λx) = 0 for any λ ∈ K ∗ .Thus if M = Kx ⊂ K n+1 is a line in K n+1 , we may say that F (M ) = 0 ifF (x) = 0, and this definition is independent of the choice of a generator of M .Now if M is a local line and Mfi = xi Kfi ⊂ Kfn+1 i for some trivializing family{fi }i∈I , we say that F (M ) = 0 if F (xi ) = 0 for all i ∈ I. This definition isindependent of the choice of a trivializing family follows from (2) above and thefollowing.Lemma 5.11. Let {fi }i∈I be a covering family in a ring A and let a ∈ A.Assume that the image of a in each Afi is equal to 0. Then a = 0.Proof. By definition of Afi , we have a/1 = 0 in Afi ⇐⇒ fin ai = 0 P for somen ≥ 0. Obviously, we choose n to be the same for all i ∈ I. Since 1 = i∈I ai fi P ai ∈ A, after raising the both sidesfor some Pin sufficient high power, we obtain1 = i∈I bi fin for some bi ∈ A. Then a = i∈I bi fin a = 0.40 LECTURE 5. PROJECTIVE ALGEBRAIC VARIETIES

F (T1 /T0 , . . . , Tn /T0 ) = T0−d G(T0 , . . . , Tn ),

G(T0 , . . . , Tn ) = T0d F (T1 /T0 , . . . , Tn /T0 )

is said to be the homogenization of F. For example, the polynomial T22 T0 + T13 +

T1 T02 + T03 is equal to the homogenization of the polynomial Z22 + Z13 + Z1 + 1. Let I be an ideal in k[Z1 , . . . , Zn ]. We define the homogenization of I asthe ideal I hom in k[T0 , . . . , Tn ] generated by homogenizations of elements of I.It is easy to see that if I = (G) is principal, then I hom = (F ), where F is thehomogenization of G. However, in general it is not true that I hom is generatedby the homogenizations of generators of I (see Problem 6 below). Recalling the injective map α0 : Ank → Pnk defined in the beginning of thislecture, we see that it sends an affine algebraic subvariety X defined by an idealI to the projective variety defined by the homogenization I hom , which is said tobe the projective closure of X. 41

of the projective plane P2k . It is equal to the projective closure of the line L ⊂ A2kgiven by the equation bZ1 + cZ2 + a = 0. For every K the set X(K) has aunique point P not in the image of L(K). Its homogeneous coordinates are(0, c, −b). Thus, X has to be viewed as L ∪ {P }. Of course, there are manyways to obtain a projective variety as a projective closure of an affine variety. Tosee this, it is sufficient to replace the map α0 in the above constructions by themaps αi , i 6= 0. Let {F (T ) = 0}F ∈S be a homogeneous system. We denote by (S) the idealin k[T ] generated by the polynomials F ∈ S. It is easy to see that this ideal hasthe following property (S) = ⊕d≥0 ((S) ∩ k[T ]d ).In other words, each polynomial F ∈ (S) can be written uniquely as a linearcombination of homogeneous polynomials from (S).Definition 5.6. An ideal I ⊂ k[T ] is said to be homogeneous if one of thefollowing conditions is satisfied: (i) I is generated by homogeneous polynomials;

(ii) I = ⊕d≥0 (I ∩ k[T ]d ).

Let us show the equivalence P of these two properties. If (i) holds, then everyF ∈ I can be written as i Qi Fi , where Fi is a set of homogeneous generators.Writing each Qi as a sum of homogeneous polynomials, we see that F is a linearcombination of homogeneous polynomials from I. This proves (ii). Assume (ii)holds. Let G1 , . . . , Gr be a system of generators of I. Writing each Gi as asum of homogeneous polynomials Gij from I, we verify that the set {Gij } is asystem of homogeneous generators of I. This shows (i). We know that in the affine case the ideal I(X) determines uniquely an affinealgebraic variety X. This is not true anymore in the projective case.Proposition 5.13. Let {F (T ) = 0}F ∈S be a homogeneous system of algebraicequations over a field k. Then the following properties are equivalent: (i) PSol(S; K)0 = ∅ for some algebraically closed field K; P (ii) (S) ⊃ k[T ]≥r := d≥r k[T ]d for some r ≥ 0;

(iii) for all k-algebras K, PSol(S; K) = ∅.

42 LECTURE 5. PROJECTIVE ALGEBRAIC VARIETIES

Proof. (i) =⇒ (ii) Let K be an algebraically closed field containing k. We can

Note that k[T ]≥r is an ideal in k[T ] which is equal to the power mr+ where m+ = k[T ]≥1 = (T0 , . . . , Tn ).A homogeneous ideal I ⊂ k[T ] containing some power of m+ is said to beirrelevant. The previous proposition explains this definition. For every homogeneous ideal I in k[T ] we define the projective algebraicvariety P V (I) as a correspondence K → PSol(I, K). We define the saturationof I by I sat = {F ∈ k[T ] : GF ∈ I for all G ∈ ms+ for some s ≥ 0}. 43

Clearly I sat is a homogeneous ideal in k[T ] containing the ideal I (Check it !) .

Proposition 5.14. Two homogeneous systems S and S 0 define the same pro-jective variety if and only if (S)sat = (S 0 )sat .

Definition 5.7. A homogeneous ideal I ⊂ k[T ] is said to be saturated if I = I sat .

Corollary 5.15. The map I → P V (I) is a bijection between the set of saturatedhomogeneous ideals in k[T] and the set of projective algebraic subvarieties of Pnk .44 LECTURE 5. PROJECTIVE ALGEBRAIC VARIETIES

In future we will always assume that a projective variety X is given by a

system of equations S such that the ideal (S) is saturated. Then I = (S) isdefined uniquely and is called the homogeneous ideal of X and is denoted byI(X). The corresponding factor-algebra k[T ]/I(X) is denoted by k[X] and iscalled the projective coordinate algebra of X. The notion of a projective algebraic k-set is defined similarly to the notionof an affine algebraic k-set. We fix an algebraically closed extension K of kand consider subsets V ⊂ Pn (K) of the form PSol(S; K), where X is a systemof homogeneous equations in n-variables with coefficients in k. We define theZariski k-topology in Pn (K) by choosing closed sets to be projective algebraick-sets. We leave the verification of the axioms to the reader.

Problems.1*. Show that Pn (k[T1 , . . . , Tn ]) = Pn (k[T1 , . . . , Tn ])0 , where k is a field.2. Let A = Z/(6). Show that A has two maximal ideals m with the correspondinglocalizations Am isomorphic to Z/(2) and Z/(3). Show that a projective A-modules of rank 1 is isomorphic to A.3*. Let A = C[T1 , T2 ]/(T12 − T2 (T2 − 1)(T2 − 2)), t1 and t2 be the cosets of theunknowns T1 and T2 . Show that the ideal (t1 , t2 ) is a projective A-module ofrank 1 but not free.4. Let I ⊂ k[T ] be a homogeneous ideal such that I ⊃ ms+ for some s. Provethat I sat = k[T ]. Deduce from this another proof of Proposition 5.13.5. Find I sat , where I = (T02 , T0 T1 ) ⊂ k[T0 , T1 ].6. Find the projective closure in P3k of an affine variety in A3k given by theequations Z2 − Z12 = 0, Z3 − Z13 = 0.7. Let F ∈ k[T0 , . . . , Tn ] be a homogeneous polynomial free of multiple factors.Show that its set of solutions in Pn (K), where K is an algebraically closedextension of k, is irreducible in the Zariski topology if and only F is an irreduciblepolynomial.Lecture 6

Bézout theorem and a group law

on a plane cubic curve

We begin with an example. Consider two ”concentric circles”:

C : Z12 + Z22 = 1, C 0 : Z12 + Z22 = 4.Obviously, they have no common points in the affine plane A2 (K) no matter inwhich algebra K we consider our points. However, they do have common points”at infinity”. The precise meaning of this is the following. Let C̄ : T12 + T22 − T02 = 0, C̄ 0 : T12 + T22 − 4T02 = 0be the projective closures of these conics in the projective plane P2k , obtained √ bythe homogenization of the corresponding polynomials. Assume √ that −1 ∈ K.Then the points (one point if K is of characteristic 2) (1, ± −1, 0) are thecommon points of C̄(K) and C̄(K)0 . In fact, the homogeneous ideal generatedby the polynomials T12 + T22 − T02 and T12 + T22 − 4T02 defining the intersectionis equal to the ideal generated by the polynomials T12 + T22 − T02 and T02 . Thesame points are the common points of the line L : T0 = 0 and the conic C̄, butin our case, it is natural to consider the same points with multiplicity 2 (becauseof T02 instead of T0 ). Thus the two conics have in some sense 4 common points.Bézout’s theorem asserts that any two projective subvarieties of P2k given by anirreducible homogeneous equation of degree m and n, respectively, have mncommon points (counting with appropriate multiplicities) in P2k (K) for everyalgebraically closed field K containing k. The proof of this theorem which weare giving here is based on the notion of the resultant (or the eliminant) of twopolynomials.

A solution can be found if and only if its determinant is equal to zero.

Obviously, this determinant is equal (up to a sign) to the value of Rn,m at(a0 , . . . , an , b0 , . . . , bm ). Conversely, assume that the above determinant van-ishes. Then we find a polynomial P1 (Z) of degree ≤ n − 1 and a polynomialQ1 (Z) of degree ≤ m−1 satisfying (1). Both of them have coefficients in k. Letα be a root of P (Z) in some extension K of k. Then α is a root of Q(Z)P1 (Z).This implies that Z − α divides Q(Z) or P1 (Z). If it divides P (Z), we founda common root of P (Z) and Q(Z). If it divides P1 (Z), we replace P1 (Z) withP1 (Z)/(Z − α) and repeat the argument. Since P1 (Z) is of degree less than n,we finally find a common root of p(Z) and q(Z).

The polynomial Rn,m is called the resultant of order (n, m). For any twopolynomials P (Z) = a0 Z n + . . . + an and Q(Z) = b0 Z m + . . . + bm the value ofRn,m at (a0 , . . . , an , b0 , . . . , bm ) is called the resultant of P (Z) and Q(Z), andis denoted by Rn,m (P, Q). A projective algebraic subvariety X of P2k given by an equation: F (T0 , T1 , T2 ) =0, where F 6= 0 is a homogeneous polynomial of degree d will be called a planeprojective curve of degree d. If d = 1, we call it a line, if d = 2 , we call ita plane conic (then plane projective curve!cubic, plane quartic, plane quintic,plane sextic and so on. We say that X is irreducible if its equation is given byan irreducible polynomial.

Theorem 6.2. (Bézout). Let

F (T0 , T1 , T2 ) = 0, G(T0 , T1 , T2 ) = 0

be two different plane irreducible projective curves of degree n and m, respec-

tively, over a field k. For any algebraically closed field K containing k, the systemF = 0, G = 0 has exactly mn solutions in P2 (K) counted with appropriate mul-tiplicities.

Proof. Since we are interested in solutions in an algebraically closed field K, we

may replace k by its algebraic closure to assume that k is algebraically closed. Inparticular k is an infinite set. We shall deduce later from the theory of dimensionof algebraic varieties that there are only finitely many K-solutions of F = G = 0.Thus we can always find a line T0 + bT1 + cT2 = 0 with coefficients in k thathas no K-solutions of F = G = 0. This is where we use the assumption thatk is infinite. Also choose a different line aT0 + T1 + dT2 = 0 with a 6= b suchthat for any λ, µ ∈ K the line (λ + µa)T0 + (λb + µ)T1 + (λc + µ)T2 = 0 has48LECTURE 6. BÉZOUT THEOREM AND A GROUP LAW ON A PLANE CUBIC CURVE

at most one solution of F = G = 0 in K. The set of triples (α, β, γ) such that

the line αT0 + βT1 + γT2 = 0 contains a given point (resp. two distinct points)is a two-dimensional (resp. one-dimensional) linear subspace of k 3 . Thus the setof lines αT0 + βT1 + γT2 = 0 containing at least two solutions of F = G = 0is a finite set. Thus we can always choose a line in k 3 containing (1, b, c) andsome other vector (a, 1, d) such that it does not belong to this set. Making theinvertible change of variables

T0 → T0 + bT1 + cT2 , T1 → aT0 + T1 + dT2 , T2 → T2

we may assume that for every solution (a0 , a1 , a2 ) of F = G = 0 we have a0 6= 0,

and also that no line of the form αT0 + βT1 = 0 contains more than one solutionof F = G = 0 in K. Write

be the corresponding homogeneous polynomial in T0 , T1 . It is easy to see, using

the definition of the determinant, that R̄ is a homogeneous polynomial of degreemn. It is not zero, since otherwise, by the previous Lemma, for every (β0 , β1 )the polynomials F (β0 , β1 , T2 ) and G(β0 , β1 , T2 ) have a common root in K. Thisshows that P2 (K) contains infinitely many solutions of the equations F = G = 0,which is impossible as we have explained earlier. Thus we may assume thatR̄ 6= 0. Dehomogenizing it, we obtain:

R̄ = T0nm R̄0 (T1 /T0 )

where R̄0 is a polynomial of degree ≤ nm in the unknown Z = T1 /T0 . Assume

Thus R̄ is obtained by ”eliminating” the unknown T2 . We see that the line

L : G = 0 “intersects” the curve X : F = 0 at n K-points corresponding ton solutions of the equation R̄(T0 , T1 ) = 0 in P1 (K). A solution is multiple, ifthe corresponding root of the dehomogenized equation is multiple. Thus we canspeak about the multiplicity of a common K-point of L and F = 0 in P2 (K).We say that a point x ∈ X(K) is a nonsingular point if there exists at most oneline L over K which intersects X at x with multiplicity > 1. A curve such thatall its points are nonsingular is called nonsingular. We say that L is tangent tothe curve X at a nonsingular point x ∈ P2 (K) if x ∈ L(K) ∩ X(K) and itsmultiplicity ≥ 2. We say that a tangent line L is an inflection tangent line at xif the multiplicity ≥ 3. If such a tangent line exists at a point x, we say that xis an inflection point (or a flex point) of X.

∂Pcoefficients in a field k. We define the partial derivatives ∂Z j of Z as follows.First we assume that P is a monomial Z1i1 · · · Znin and set ( i −1 ∂P ij Z1i1 · · · Zjj · · · Znin if ij > 0, = . ∂Zj 0 otherwise50LECTURE 6. BÉZOUT THEOREM AND A GROUP LAW ON A PLANE CUBIC CURVE

Then we extend the definition to all polynomials by linearity over k requiring that

∂(aP + bQ) ∂P ∂Q =a +b ∂Zj ∂Zj ∂Zj

for all a, b ∈ k and any monomials P, Q. It is easy to check that the partialderivatives enjoy the same properties as the partial derivatives of functions defined ∂Pby using the limits. For example, the map P 7→ ∂Z j is a derivation of the k-algebra k[Z1 , . . . , Zn ], i.e. , it is a k-linear map ∂ satisfying the chain rule:

∂(P Q) = P ∂(Q) + Q∂(P ).

The partial derivatives of higher order are defined by composing the operators ofpartial derivatives.

Proof. We check these assertions only for the case (a0 , a1 , a2 ) = (1, 0, 0). Thegeneral case is reduced to this case by using the variable change. The usualformula for the variable change in partial derivatives are easily extended to our 51

= 2(d − 1)2 P2 (a, b).

It follows from (6.2) that P2 (a, b) = 0 if and only if P2 (1, λ) = 0. Since weassume that (char(k), d − 1) = 1, we obtain that (1, 0, 0) is an inflection pointif and only if the determinant from assertion (iii) is equal to zero.52LECTURE 6. BÉZOUT THEOREM AND A GROUP LAW ON A PLANE CUBIC CURVE

Remark 6.5. The determinant

is a homogeneous polynomial of degree 3(d − 2) unless it is identically zero. It is

called the Hessian polynomial of F and is denoted by Hess(F ). If Hess(F ) 6= 0,the plane projective curve of degree 3(d − 2) given by the equation Hess(F ) = 0is called the Hessian curveindexHessian curve of the curve F = 0. ApplyingProposition 6.4 and Bézout’s Theorem, we obtain that a plane curve of degreed has 3d(d − 2) inflection points counting with multiplicities. Here is an example of a polynomial F defining a nonsingular plane curve withHess(F ) = 0:

F (T0 , T1 , T2 ) = T0p+1 + T1p+1 + T2p+1 = 0,

where k is of characteristic p > 0. One can show that Hess(F ) 6= 0 if k is of

characteristic 0. Let us give an application of the Bézout Theorem. Let

algebraically closed). Let k̄ be the algebraic closure of k containing K. Weassume that each point of X(k̄) is nonsingular. Later when we shall study localproperties of algebraic varieties, we give some simple criterions when does ithappen. Fix a point e ∈ X(K). Let x, y be two different points from X(K). Definethe sum x ⊕ y ∈ X(K)as a point in X(K) determined by the following construction. Find a line L1over K with y, x ∈ L1 (K). This can be done by solving two linear equationswith three unknowns. By Bézout’s Theorem, there is a third intersection point,denote it by yx. Since this point can be found by solving a cubic equation overK with two roots in K (defined by the points x and y), the point yx ∈ X(K).Now find another K-line L2 which contains yx and e, and let y ⊕ x denote thethird intersection point. If yx happens to be equal to e, take for L2 the tangent 53

x y xy ox⊕y

Figure 6.1:

line to X at e. If y = x, take for L1 the tangent line at y. We claim that this

construction defines the group law on X(K). Clearly y ⊕ x = x ⊕ y,i.e., the binary law is commutative. The point e is the zero element of the law. Ifx ∈ X(K), the opposite point −x is the point of intersection of X(K) with theline passing through x and the third point x1 at which the tangent at e intersectsthe curve. The only non-trivial statement is the property of associativity. Consider the eight points e, x, y, z, zy, xy, x ⊕ y, y ⊕ z. They lie on threecubic curves. The first one is the original cubic X. The second one is the unionof three lines < x, y > ∪ < yz, y ⊕ z > ∪ < z, x ⊕ y > (6.3)where for any two distinct points a, b ∈ P2 (K) we denote by < a, b > the uniqueK-line L with a, b ∈ L(K). Also the “union” means that we are considering thevariety given by the product of the linear polynomials defining each line. Thethird one is also the union of three lines

< y, z > ∪ < xy, x ⊕ y > ∪ < x, y ⊕ z > . (6.4)

We will use the following:

Lemma 6.6. Let x1 , . . . , x8 be eight distinct points in P2 (K). Suppose that allof them belong to X(K) where X is a plane irreducible projective cubic curve.Assume also that the points x1 , x2 , x3 lie on two different lines which do notcontain points xi with i > 3. There exists a unique point x9 such that any cubiccurve Y containing all eight points contains also x9 , and either x9 6∈ {x1 , . . . , x8 }or x9 enters in X(K) ∩ Y (K) with multiplicity 2.54LECTURE 6. BÉZOUT THEOREM AND A GROUP LAW ON A PLANE CUBIC CURVE

polynomial F . A point x = (α0 , α1 , α2 ) ∈ X(K) if and only if the ten coefficientsof F satisfy a linear equation whose coefficients are the values of the monomialsof degree 3 at (α0 , α1 , α2 ). The condition that a cubic curve passes through8 points introduces 8 linear equations in 10 unknowns. The space of solutionsof this system is of dimension ≥ 2. Suppose that the dimension is exactly 2.Then the equation of any cubic containing the points x1 , . . . , x8 can be writtenin the form λF1 + µF2 , where F1 and F2 correspond to two linearly independentsolutions of the system. Let x9 be the ninth intersection point of F1 = 0and F2 = 0 (Bézout’s Theorem). Obviously, x9 is a solution of F = 0. Itremains to consider the case when the space of solutions of the system of linearequation has dimension > 2. Let L be the line with x1 , x2 ∈ L(K). Choose twopoints x, y ∈ L(K) \ {x1 , x2 } which are not in X(K). Since passing througha point imposes one linear condition, we can find a cubic curve Y : G = 0with x, y, x1 , . . . , x8 ∈ Y (K). But then L(K) ∩ Y (K) contains four points.By Bézout’s Theorem this could happen only if G is the product of a linearpolynomial defining L and a polynomial B of degree 2. By assumption L doesnot contain any other point x3 , . . . , x8 . Then the conic C : B = 0 must containthe points x3 , . . . , x8 . Repeating the argument for the points x1 , x3 , we find aconic C 0 : B 0 = 0 which contains the points x2 , x4 , . . . , x8 . Clearly C 6= C 0since otherwise C contains 7 common points with an irreducible cubic. SinceC(K) ∩ C 0 (K) contains 5 points in common, by Bézout’s Theorem we obtainthat B and B 0 have a common linear factor. This easily implies that 4 pointsamong x4 , . . . , x8 must be on a line. But this line cannot intersect an irreduciblecubic at four points in P2k (K).

Here is an example of the configuration of 8 points which do not satisfy

the assumption of Lemma 6.6. Consider the cubic curve (over C) given by theequation: T03 + T13 + T23 + λT0 T1 T2 = 0.It is possible to choose the parameter λ such that the curve is irreducible. Letx1 , . . . , x9 be the nine points on this curve with the coordinates:

(0, 1, ρ), (1, 0, ρ), (1, 1, ρ)

where ρ is one of three cube roots of −1. Each point xi lies on four lineswhich contain two other points xj 6= xi . For example, (0, 1, −1) lies on theline T0 = 0 which contains the points (0, 1, ρ), (0, 1, ρ2 ) and on the three lines 55

Nevertheless one can prove that the assertion of Lemma 6.6 is true withoutadditional assumption on the eight points. To apply Lemma 6.6 we take the eight points e, x, y, z, zy, xy, x ⊕ y, y ⊕ zin X(K). Obviously, they satisfy the assumptions of the lemma. Observe that(x ⊕ y)z lies in X(K) and also in the cubic (6.3), and x(y ⊕ z) lies in X(K)and in the cubic (6.4). By the Lemma (x ⊕ y)z = x(y ⊕ z) is the unique ninthpoint. This immediately implies that (x ⊕ y) ⊕ z = x ⊕ (y ⊕ z).Remark 6.7. Our proof is in fact not quite complete since we assumed that all thepoints e, x, y, z, zy, xy, x⊕y, y⊕z are distinct. We shall complete it later but theidea is simple. We will be able to consider the product X(K) × X(K) × X(K)as a projective algebraic set with the Zariski topology. The subset of triples(x, y, z) for which the associativity x ⊕ (y ⊕ z) = (x ⊕ y) ⊕ z holds is open(since all degenerations are described by algebraic equations). On the otherhand it is also closed since the group law is defined by a polynomial map. Since56LECTURE 6. BÉZOUT THEOREM AND A GROUP LAW ON A PLANE CUBIC CURVE

X(K) × X(K) × X(K) is an irreducible space, this open space must coincidewith the whole space.Remark 6.8. Depending on K the structure of the group X(K) can be verydifferent. A famous theorem of Mordell-Weil says that this group is finitelygenerated if K is a finite extension of Q. One of the most interesting problemsin number theory is to compute the rank of this group. On the other hand, thegroup X(C) is isomorphic to the factor group C/Z2 . Obviously, it is not finitelygenerated.

3. Find explicit formulae for the group law on X(C), where X is a cubic curvedefined by the equation T12 T0 − T23 − T03 = 0. You may take for the zero element thepoint (0, 1, 0). 4. In the notation of the previous problem, show that elements x ∈ X(C) of order3 (i.e. 3x = 0 in the group law) correspond to inflection points of X. Show thatthere are 9 of them. Show that the set of eight inflection points is an example of theconfiguration which does not satisfy the assumption of Lemma 6.6. 5. Let X be given by the equation T12 T0 − T23 = 0. Similarly to the case of anonsingular cubic, show that for any field K the set X(K)0 = X(K) \ {(1, 0, 0)} hasa group structure isomorphic to the additive group K + of the field K. 6. Let X be given by the equation T12 T0 − T22 (T2 + T0 ) = 0. Similarly to the caseof a nonsingular cubic, show that for any field K the set X(K)0 = X(K) \ {(1, 0, 0)}has a group structure isomorphic to the multiplicative group K ∗ of the field K.Lecture 7

Morphisms of projective algebraic

varieties

Following the definition of a morphism of affine algebraic varieties we can define a

is commutative. Recall that a morphism of affine varieties f : X → Y is uniquely

determined by the homomorphism f ∗ : O(Y ) → O(X). This is not true anymore forprojective algebraic varieties. Indeed, let φ : k[Y ] → k[X] be a homomorphism of theprojective coordinate rings. Suppose it is given by the polynomials F0 , . . . , Fn . Thenthe restriction of the map to the set of global lines must be given by the formula a = (α0 , . . . , αn ) → (F0 (a), . . . , Fn (a)).Obviously, these polynomials must be homogeneous of the same degree. Otherwise,the value will depend on the choice of coordinates of the point a ∈ X(K). This isnot all. Suppose all Fi vanish at a. Since (0, . . . , 0) 6∈ C(K)n , the image of a is notdefined. So not any homomorphism k[Y ] → k[X] defines a morphism of projectivealgebraic varieties. In this lecture we give an explicit description for morphisms ofprojective algebraic varieties. Let us first learn how to define a morphism f : X → Y ⊂ Pnk from an affinek-variety X to a projective algebraic k-variety Y . To define f it is enough to define

where gij ∈ O(X)∗ai bj .

For each i ∈ I this collection defines a projective module Mi ∈ Pn (O(X)ai ) (i) (i)generated by (p0 , . . . , pn ). We shall prove in the next lemma that there existsa projective module M ∈ Pn (O(X)) such that Mai ∼ = Mi for each i ∈ I. Thismodule is defined uniquely up to isomorphism. Using M we can define f by sendingidO(X) ∈ X(O(X)) to M . If x ∈ X(K), where K is a field, the image fK (x) isdefined by formulae (7.3).60 LECTURE 7. MORPHISMS OF PROJECTIVE ALGEBRAIC VARIETIES

Let us now state and prove the lemma. Recall first that for any ring A a local lineM ∈ Pn (A) defines a collection {Mai }i∈I of lines in An+1 ai for some covering family{ai }i∈I of elements in A. Let us see how to reconstruct M from {Mai }i∈I . We knowthat for any i, j ∈ I the images mi of m ∈ M in Mai satisfy the following conditionof compatibility: ρij (mi ) = ρji (mj )where ρij : Mai → Mai fj is the canonical homomorphism m/ari → mfjr /(ai fj )r . For any family {Mi }i∈I of Aai -modules let Y limindi∈I Mi = {(mi )∈I ∈ Mi : ρij (mi ) = ρji (mj ) for any i, j ∈ I}. i∈I QThis can be naturally considered as a submodule of the direct product i∈I Mi ofA-modules. There is a canonical homomorphism

α : M → limindi∈I Mai

defined by m → (mi = m)i∈I .

Lemma 7.1. The homomorphism

α : M → limindi∈I Mai

is an isomorphism.

Proof. We assume that the set of indices I is finite. This is enough for our applicationssince we can always choose a finite covering subfamily. The proof of injectivity is similarto the proof of Lemma 5.11 and is left to the reader. Let us show the surjectivity. Let mi ( )i∈I ∈ limindi∈I Mai ani i

This shows that the image of m in each Mai coincides with pi /aki = mi /ani for eachi ∈ I. This proves the surjectivity. (i) (i) In our situation, Mi is generated by (p0 , . . . , pn ) ∈ C(O(Xai ) and property (ii)from above tells us that (Mi )aj = (Mj )ai . Thus we can apply the lemma to defineM. Let f : X → Y be a morphism of projective algebraic varieties, X ⊂ Pm n k , Y ⊂ Pk .For every k-algebra K and M ∈ X(K) we have N = fK (M ) ∈ Y (K). It followsfrom commutativity of diagrams (7.1) that for any a ∈ K, f (Ka )(Ma ) = Na . Let{ai }i∈I be a covering family of elements in K. Then, applying the previous lemma,we will be able to recover N from the family {Nai }i∈I . Taking a covering family whichtrivializes M , we see that our morphism f : X → Y is determined by its restrictionto X 0 : K → Pn (K)0 ∩ X(K), i.e., it suffices to describe it only on ”global” linesM ∈ X(K). Also observe that we can always choose a trivializing family {ai }i∈I ofany local line M ∈ X(K) in such a way that Mai is given by projective coordinates (i) (i) (i)(t0 , . . . , tm ) with at least one tj invertible in Aai . For example we can take the (i) (i)covering family, where each ai is replaced by {ai t0 , . . . , ai tm } (check that it is a (i)covering family) then each tj is invertible in Ka t(i) . Note that this is true even when i j (i)tj = 0 because K0 = {0} and in the ring {0} one has 0 = 1. Thus it is enoughto define the maps X(K) → Y (K) on the subsets X(K)00 of global K-lines with atleast one invertible projective coordinate. Let X be defined by a homogeneous ideal I ⊂ k[T0 , . . . , Tm ]. We denote by Ir theideal in the ring k[T0 /Tr , . . . , Tm /Tr ] obtained by dehomogenizations of polynomialsfrom I. Let Xr ⊂ Am k be the corresponding affine algebraic k-variety. We have ∼O(Xr ) = k[T0 /Tr , . . . , Tm /Tr ]/Ir . We have a natural map ir : Xr (K) → X(K)00obtained by the restriction of the natural inclusion map ir : K m → Pm (K)00 (putting1 at the rth spot). It is clear that each x ∈ X(K)00 belongs to the image of someir . Now to define the morphism X → Y it suffices to define the morphisms fr :Xr → Y, r = 0, . . . , m. This we know how to do. Each fr is given by a collection (s) (s) (s){(p0 , . . . , pn )}s∈S(r) , where each coordinate pj is an element of the ring O(X)r ), (s) (s)and as ∈ rad({p0 , . . . , pn }) for some as ∈ O(X)r . We can find a representative (s) (s) d (s)of pj in k[T0 /Tr , . . . , Tm /Tr ] of the form Pj /Tr j where Pj is a homogeneouspolynomials of the same degree dj . Reducing to the common denominator, we canassume that dj = d(s) is independent of j = 0, . . . , n. Also by choosing appropriate62 LECTURE 7. MORPHISMS OF PROJECTIVE ALGEBRAIC VARIETIES

(a0 , a1 , a2 ) → (a0 , a2 ) → (a0 a2 , a20 , a22 ) =

Similarly, we check that the other composition of the functor morphisms is the identity.Recall that the affine circle X is not isomorphic to the affine line A1k .

Example 7.5. A projective subvariety E of Pnk is said to be a projective subspace of

dimension d (or d-flat) if it is given by a system of linear homogeneous equations withcoefficients in k, whose set of solutions in k n+1 is a linear subspace E of k n+1 ofdimension d + 1. It follows from linear algebra that each such E can be given by ahomogeneous system of linear equations

Example 7.6. We already know that P1k is isomorphic to a subvariety of P2k given by n+man equation of degree 2. This result can be generalized as follows. Let N = m −1.Let us denote the projective coordinates in PNk by

Ti = Ti0 ...in , i0 + . . . + in = |i| = m.

Choose some order in the set of multi-indices i with |i| = m. Consider the morphism(the Veronese morphism of degree m)

With this definition in mind we can say that the correspondence K → Pn (K) ×Pm (K) is a projective algebraic variety. We leave to the reader to define the notions of a morphism and isomorphismbetween projective algebraic k-varieties. For example, one defines the projection morphisms:

where the symmetric matrix (aij ) is nonsingular, is isomorphic to P1k × P1k . Give anexplicit formula for the projection maps: pi : Q → P1k .3. Show that Vern1 is isomorphic to the projective closure of the affine curve given bythe equations {Zn − Z1n = 0, . . . , Z2 − Z12 = 0} (a rational normal curve of degree n).Compare this with the Problem 6 of Lecture 5.4. Show that the image of a linear projection of the twisted cubic curve in P3k from apoint not lying on this curve is isomorphic to a plane cubic curve. Find its equationand show that this curve is singular in the sense of the previous lecture.5. Show that the symmetric m-power S m (M ) of a projective module is a projectivemodule. Using this prove that the Veronese map vn,m is defined by the formulaM → Symm (M ).6. a) Show that Pn (K)00 × Pm 00 k (K) is naturally bijectively equivalent to the set of(n + 1) × (n + 1) matrices of rank 1 with coefficients in K defined up to multiplicationby a nonzero scalar.b) Show that Ver2n (K)00 is naturally bijectively equivalent to the set of symmetric rank1 square matrices of size n + 1 with coefficients in K defined up to multiplication bya nonzero scalar.7. Construct a morphism from P1k to the curve X equal to the projective closure ofthe affine curve (Z12 + Z22 )2 − Z2 (3Z12 − Z22 )) ⊂ A2k . Is X isomorphic to P1k ?Lecture 8

Quasi-projective algebraic sets

Let k be a field and K be an algebraically closed field containing k as a subfield.

Definition 8.1. A projective algebraic set over k (or a projective algebraic k-set) is asubset V of Pn (K) such that there exists a projective algebraic variety X over k withX(K) = V .

The variety X with X(K) = V 6= ∅ is not defined uniquely by V . However, as

follows from the Nullstellensatz

X(K) = Y (K) ⇐⇒ rad(I(X)) = rad(I(Y )).

Thus, if we require that X is given by a radical homogeneous ideal, the variety X is

determined uniquely by the set X(K). In the following we will always assume this.Note that a radical homogeneous ideal I coincides with its saturation I sat . Indeed, ifms F ∈ I for some s and F ∈ k[T ]d then all monomials entering into F belong to md .In particular, F s ∈ mds ⊂ ms and F s F = F s+1 ∈ I. Since I is radical this impliesthat F ∈ I. In fact we have shown that, for any ideal I, we have

I ⊂ I sat ⊂ rad(I).

This, if I = rad(I), then I = I sat . Since a projective algebraic k-variety is uniquely

determined by a saturated homogeneous ideal, we see that there is a bijective cor-respondence between projective algebraic k-sets and projective algebraic k-varietiesdefined by a radical homogeneous ideal (they are called reduced projective algebraick-varieties). We can consider Pn (K) as a projective algebraic set over any subfield k of K.Any projective algebraic k-subset of Pn (K) is called a closed subset of Pn (K). Thereason for this definition is explained by the following lemma.

6970 LECTURE 8. QUASI-PROJECTIVE ALGEBRAIC SETS

Proposition 8.1. There exists a unique topology on the set Pn (K) whose closedsubsets are projective algebraic k-subsets of Pnk (K).

Proof. This is proven similarly to that in the affine case and we omit the proof.

The topology on Pn (K) whose closed sets are projective algebraic subsets is saidto be the Zariski k-topology. We will denote the corresponding topological space byPnk (K). As is in the affine case we will drop k from the definitions and the notationsif k = K. Every subset of Pnk (K) will be considered as a topological subspace withrespect to the induced Zariski k-topology.

Lemma-Definition 8.2. A subset V of a topological space X is said to be locally

closed if one the following equivalent properties holds:

(i) V = U ∩ Z, where U is open and Z is closed;

(ii) V is an open subset of a closed subset of X;

(iii) V = Z1 \ Z2 , where Z1 and Z2 are closed subsets of X.

Proof. Left to the reader.

Definition 8.2. A locally closed subset subset of Pnk (K) is called a quasi-projectivealgebraic k-set.

In other words, a quasi-projective k-subset of Pn (K) is obtained by taking the set

of K-solutions of a homogeneous system of algebraic equations over k and throwingaway a subset of the solutions satisfying some additional equations. An example of an open quasi-projective subset is the subset

Pn (K)i = {(a0 , . . . , an ) ∈ Pn (K) : ai 6= 0}.

Its complement is the “coordinate hyperplane”:

Hi = {(a0 , . . . , an ) ∈ Pn (K) : ai = 0}.

Every affine algebraic k-set V ⊂ Ank (K) can be naturally considered as a quasi-projective algebraic set. We view An (K) = K n as the open subset Pn (K)0 , then notethat V = V̄ ∩ Pn (K)0 , where V̄ is the closure of V defined by the homogenization ofthe ideal defining V . It is clear that, in general V is neither open nor closed subsetof Pn (K). Also observe that V̄ equals the closure in the sense of topology, i.e., theminimal closed subset of Pnk (K) which contains V . Next, we want to define regular maps between quasi-projective algebraic sets. 71

called regular if there exists a finite open cover V = ∪i Ui such that the restriction off to each open subset Ui is given by a formula: (i) (i) x → (F0 (x), . . . , Fm (x)), (i) (i)where F0 (T ), . . . , Fm (T ) are homogeneous polynomials of some degree di with co-efficients in k.

Proof. We have shown in Lecture 7 that the restriction of fK to each open set V ∩(Pn )iis given by several collections of homogeneous polynomials. Each collection is definedon an open set of points where some element of a covering family does not vanish.

f (x) = (F0 (x), F1 (x)) = (1, F1 (x)/F0 (x))

can be identified with the element F1 (x)/F0 (x) of the field K = A1 (K). Thus fis given in Ui by a function of the form F/G, where F and G are homogeneouspolynomials of the same degree with G(x) 6= 0 for all x ∈ Ui . Two such functionsF/G and F 0 /G0 are equal on Ui if and only if (F G0 − F 0 G)(x) = 0 for all x ∈ Ui . IfV is irreducible this implies that (F G0 − F 0 G)(x) = 0 for all x ∈ V . A regular map f : V → A1 (K) is called a regular function on V . The set of regularfunctions form a k- algebra with respect to multiplication and addition of functions.We shall denote it by O(V ). As we will prove later O(V ) = k if V is a projectivealgebraic k-set. On the opposite side we have:

Proposition 8.4. Let V ⊂ An be an affine algebraic set considered as a closed subset

in Pn (K)i . Then O(V ) is isomorphic to the algebra of regular function of the affinealgebraic set V .

This shows that f is a global polynomial map, i.e. a regular function on V .

An isomorphism (or a biregular map) of quasi-projective algebraic sets is a bijective

regular map such that the inverse map is regular (see Remark 3.7 in Lecture 3 whichshows that we have to require that the inverse is a regular map). Two sets areisomorphic if there exists an isomorphism from one set to another. It is not difficult to see (see Problem 8) that a composition of regular maps is aregular map. This implies that a regular map f : V → W defines the homomorphismof k-algebras f ∗ (O(W ) → O(V ). However, in general, this homomorphism does notdetermine f uniquely (as in the case of affine algebraic k-sets).

Definition 8.4. A quasi-projective algebraic set is said to be affine if it is isomorphic

to an affine algebraic set. 73

Let V be a closed subset of Pn (K) defined by an irreducible homogeneous poly-

nomial F of degree m > 1. The complement set U = Pn (K) \ V does not come fromany closed subset of Pn (K)i since V does not contain any hyperplane Ti = 0. So,U is not affine in the way we consider any affine set as a quasi-projective algebraicset. However, U is affine. In fact, let vn,m : Pn (K) → PN (n,m) be the Veronese mapdefined by monomials of degree m. Then vn,m (U ) is contained in the complement ofa hyperplane H in PN (n,m) defined by considering F as a linear combination of mono-mials. composing vn,m with a projective linear transformation we may assume that His a coordinate hyperplane. Thus vn,m defines an isomorphism from U to the opensubset of the Veronese projective algebraic set V ern,m (K) = vn,m (Pn (K)) whosecomplement is the closed subset V ern,m (K) ∩ H. But this set is obviously affine, it isdefined in PN (n,m) (K)i = K N (n,m) by dehomogenizations of the polynomials definingVern,m .

Lemma 8.5. Let V be an affine algebraic k-set and f ∈ O(V ). Then the set

D(f ) = {x ∈ V : f (x) 6= 0}

is affine and O(D(f )) ∼ = O(V )f .

Proof. Replacing V by an isomorphic algebraic k-set, we may assume that V = X(K),

Theorem 8.6. Let V be a quasi-projective k-set and x ∈ V . Then there exists an

open subset U ⊂ V containing x which is an affine quasi-projective set.Proof. Let V = Z1 \ Z2 , where Z1 , Z2 are closed subsets of Pnk (K). Obviously,x ∈ Pn (K)i for some i. Thus x belongs to (Z1 ∩ Pn (K)i ) \ (Z2 ∩ Pn (K)i ). Thesubsets Z1 ∩ Pn (K)i and Z2 ∩ Pn (K)i are closed subsets of K n . Let F be a regularfunction on K n which vanishes on Z2 ∩ Pn (K)i but does not vanish at x. Then itsrestriction to V ∩(Z1 ∩Pn (K)i ) defines a regular function f ∈ O(V ∩Pn (K)i ) such thatx ∈ D(f ) ⊂ V ∩ Pn (K)i . By the previous lemma, D(f ) is an affine quasi-projectivek-set.Corollary 8.7. The set of open affine quasi-projective sets form a basis in the Zariskitopology of Pn (K). Recall that a basis of a topological space X is a family F of open subsets such thatfor any x ∈ X and any open U containing x there exists V ∈ F such that x ∈ V ⊂ U .We shall prove in the next lecture that the intersection of two open affine sets is anopen affine set. This implies that the Zariski topology can be reconstructed from theset of affine open sets.Remark 8.8. The reader who is familiar with the notion of a manifold (real or complex)will easily notice the importance of the previous theorem. It shows that the notion ofa quasi-projective algebraic set is very similar to the notion of a manifold. A quasi-projective algebraic set is a topological space which is locally homeomorphic to aspecial topological space, an affine algebraic set.Proposition 8.9. Every quasi-projective algebraic k-set V is a quasi-compact topo-logical space.Proof. Recall that a topological space V (not necessarily separated) is said to bequasi-compact if every its open covering {Ui }i∈I contains a finite subcovering, i.e.

V = ∪i∈I Ui =⇒ V = ∪i∈J Ui ,

where J is a finite subset of I.

Every Noetherian space is quasi-compact. Indeed, in the above notation we forma decreasing sequence of closed subsets

V \ Ui1 ⊃ V \ (Ui1 ∪ Ui2 ) ⊃ . . .

which must stabilize with a set V 0 = V \ (Ui1 ∪ . . . ∪ Uir ). If it is not empty, we cansubtract one more subset Uij to decrease V 0 . Therefore, V 0 = ∅ and V = Ui1 ∪. . .∪Uir .Thus, it suffices to show that a quasi-projective set is Noetherian. But obviously itsuffices to verify that its closure is Noetherian. This is checked similarly to that as inthe affine case by applying Hilbert’s Basis Theorem. 75

Corollary 8.10. Every algebraic set can be written uniquely as the union of finitelymany irreducible subspaces Zi , such that Zi 6⊂ Zj for any i 6= j.

Lemma 8.11. Let V be a topological space and Z be its subspace. Then Z is

irreducible if and only if its closure Z̄ is irreducible.

Proof. Obviously, follows from the definition.

Proposition 8.12. A subspace Z of Pnk (K) is irreducible if and only if the radicalhomogeneous ideal defining the closure of Z is prime.

Proof. By the previous lemma, we may assume that Z is closed. Then Z is a projectivealgebraic set defined by its radical homogeneous ideal. The assertion is proven similarlyto the analogous assertion for an affine algebraic set. We leave the proof to thereader.

The image of a projective

algebraic set

Let f : V → W be a regular map of quasi-projective k-sets. We are interested in its

image f (V ). Is it a quasi-projective algebraic set? For instance, let f : K 2 → K 2 begiven by (x, y) 7→ (x, xy). Then its image is the union of the set U = {(a, b) ∈ K 2 :a 6= 0} and the closed subset Z = {(0, 0)}. The complement of a a locally closedsubset is equal to the union of an open a closed set. If the open part is not empty,then closure must be equal to K 2 . The complement of f (K 2 ) is equal to the set ofpoints (0, y), y 6= 0. Obviously, its closure is the line y = 0 and it does not containany open subsets of K n . Thus f (A2k (K)) is not locally closed in A2k (K). Since K 2is an open subset of P2k (K), f (A2k (K)) is not locally closed in P2k (K), i.e., it is not aquasi-projective algebraic set. However, the situation is much better in the case where V is a projective set. Wewill prove the following result:Theorem 9.1. The image of a projective algebraic k-set V under a regular mapf : V → W is a closed subset of W in the Zariski k-topology. To prove this theorem we note first that

Pnk (K) × Pm n m k (K) (resp. of Pk (K) × Ak (K)) defined in this way is a closed subset inthe Zariski k-topology of the product.Proof. It is enough to prove the first statement. The second one will follow fromthe first one by taking the closure of V in Pnk (K) × Pm k (K) and then applying the 0dehomogenization process in the variable T0 . Now we know that V is given by a system (n+1)(m+1)−1of homogeneous polynomials in variables Tij in the space Pk and the systemof equations defining the Segre set Segn,m (K). Using the substitution Tij = Ti Tj0 ,we see that V can be given by a system of equations in T0 , . . . , Tn , T00 , . . . , Tm 0 which

are homogeneous in each set of variables of the same degree. If we have a systemof polynomials Ps (T0 , . . . , Tn , T00 , . . . , Tm 0 ) which are homogeneous of degree d(s) in

of solutions in Pnk (K) × Pm k (K) is also given by the system in which we replace each 0d(s)−d(s)0 d(s)0 −d(s)Ps by Ti Ps , i = 0, . . . , m, if d(s) > d(s)0 and by Ti Ps , i = 0, . . . , n,if d(s) < d(s)0 . Then the enlarged system arises from a system of polynomials in Tijafter substitution Tij = Ti Tj0 .

algebraic subset of Pn (K) defined by the system of homogeneous equations:

Fi (T0 , . . . , Tn , a1 , . . . , am ) = 0, i = 1, . . . , N.

It is clear that Xa = ∅ if and only if (0, . . . , 0) is the only solution of this system inK n+1 . By Nullstellensatz, this happens if only if the radical of the ideal Ia generated bythe polynomials Fi (T, a1 , . . . , am ) is equal to (T0 , . . . , Tn ). This of course equivalentto the property that (T0 , . . . , Tn )s ⊂ Ia for some s ≥ 0. Now we observe that

This map is surjective if and only if a ∈ K m \ Ys . Thus, a ∈ Ys if and only if

rank(φ) < d = dimk[T ]s . The latter condition can be expressed by the equality tozero of all minors of order d in any matrix representing the linear map φ. However,the coefficients of such a matrix (for example, with respect to a basis formed bymonomials) are polynomials in a1 , . . . , an with coefficients from k. Thus, every minoris also a polynomial in a. The set of zeros of these polynomials defines the closedsubset Ys in the Zariski k-topology. This proves Theorem 9.3. Recall that a topological space X is said to be connected if X 6= X1 ∪ X2 whereX1 and X2 are proper open (equivalently, closed) subsets with empty intersection.One defines naturally the notion of a connected component of V and shows that Vis the union of finitely many connected components. Clearly, an irreducible spaceis always connected, but the converse is false in general. For every quasi-projectivealgebraic k-set V we denote by π0 (V ) the set of its connected components. Let π̄0 (V )denote the set of connected components of the corresponding K-set. Both of thesesets are finite since any irreducible component of V is obviously connected. We saythat V is geometrically connected if #π̄0 (V ) = 1. Notice the difference betweenconnectedness and geometric connectedness. For example, the number of connectedcomponents of the affine algebraic k-subset of A1k defined by a non-constant non-zero polynomial F (Z) ∈ k[Z] equals the number of irreducible factors of F (Z). The 81

number of connected components of the corresponding K-set equals the number of

Corollary 9.7. Let Z be a closed connected subset of Pnk (K). Suppose Z is containedin an affine subset U of Pnk (K). Then the ideal of O(U ) of functions vanishing on Zis a maximal ideal. In particular, Z is one point if k is algebraically closed.

Proof. Obviously, Z is closed in U , hence is an affine algebraic k-set. We know that

O(Z) = k 0 is a finite field extension of k. The kernel of the restriction homomorphismresU/Z : O(U ) → O(Z) = k 0 is a maximal ideal in O(U ). In fact if A is a subringof k 0 containing k it must be a field (every nonzero x ∈ A satisfies an equationxn + a1 xn−1 + . . . + an−1 x + an = 0 with an 6= 0, hence x(xn−1 + a1 xn−2 + . . . +an−1 )(−a−1 n ) = 1). This shows that Z does not contain proper closed subsets in theZariski k-topology. If k is algebraically closed, all points are closed, hence Z must bea singleton.82 LECTURE 9. THE IMAGE OF A PROJECTIVE ALGEBRAIC SET

Proof. We may assume that k = K since we are talking about algebraic K-sets. LetW ⊂ Pn (K)0 ⊂ Pn (K) for some n, and f 0 : V → Pn (K) be the composition of fand the natural inclusion W ,→ Pn (K). By Theorem 9.1, f (V ) = f 0 (V ) is a closedconnected subset of Pn (K) contained in an affine set (the image of a connected setunder a continuous map is always connected). By Corollary 9.7, f (V ) must be asingleton.

Problems.1. Let K[T0 , . . . , Tn ]d be the space of homogeneous polynomials of degree d withcoefficients in an algebraically closed field K. Prove that the subset of reduciblepolynomials is a closed subset of K[T0 , . . . , Tn ]d where the latter is considered asaffine space AN (K), N = n+d d . Find its equation when n = d = 2.2. Prove that K n \ {a point} or Pn (K) \ {point} is not an affine algebraic set ifn > 1, also is not isomorphic to a projective algebraic set.3. Prove that the intersection of open affine subsets of a quasi-projective algebraic setis affine [Hint: Use that for any two subsets A and B of a set S, A∩B = ∆S ∩(A×B)where the diagonal ∆S is identified with S].4. Let X ⊂ Pn be a connected projective algebraic set other than a point and Y is aprojective set defined by one homogeneous polynomial. Show that X ∩ Y 6= ∅.5. Let f : X → Z and g : Y → Z be two regular maps of quasi-projective algebraicsets. Define X ×Z Y as the subset of X × Y whose points are pairs (x, y) such thatf (x) = g(y). Show that X ×Z Y is a quasi-projective set. A map f : X → Z is calledproper if for any map g : Y → Z and any closed subset W of X ×Z Y the image ofW under the second projection X × Y → Y is closed. Show that f is always properif X is a projective algebraic set.Lecture 10

Finite regular maps

The notion of a finite regular map of algebraic sets generalizes the notion of a finiteextension of fields. Recall that an extension of fields F → E is called finite if E isa finite-dimensional vector space over F . This is easy to generalize. We say that aninjective homomorphism φ : A → B of commutative rings is finite if B considered asa module over A via the homomorphism φ is finitely generated. What is the geometricmeaning of this definition? Recall that a finite extension of fields is an algebraicextension. This means that any element in E satisfies an algebraic equation withcoefficients in F . The converse is also true provided E is finitely generated over F asa field. We shall prove in the next lemma that a finite extension of rings has a similarproperty: any element in B satisfies an algebraic equation with coefficients in φ(A).Also the converse is true if we additionally require that B is a finitely generated algebraover A and every element satisfies a monic equation (i.e. with the highest coefficientequal to 1) with coefficients in φ(A). Let us explain the geometric meaning of the additional assumption that the equa-tions are monic. Recall that an algebraic extension E/F has the following property.Let y : F → K be a homomorphism of F to an algebraically closed field K. Theny extends to a homomorphism of fields x : E → K. Moreover, the number of theseextensions is finite and is equal to the separable degree [E : F ]s of the extensionE/F . An analog of this property for ring extensions must be the following. For anyalgebraically closed field K which has a structure of a A-algebra via a homomorphismy : A → K (this is our analog of an extension K/F ) there a non-empty finite set ofhomomorphisms xi : B → K such that xi ◦ φ = y. Let us interpret this geometricallyin the case when φ is a homomorphism of finitely generated k-algebras. Let X and Ybe afiine algebraic k-varieties such that O(X) ∼= B, O(Y ) ∼ = A. The homomorphism ∗φ defines a morphism f : X → Y such that φ = f . A homomorphism y : A → K isa K-point of Y . A homomorphism yi : B → K such that xi ◦ φ = y is a K-point of

8384 LECTURE 10. FINITE REGULAR MAPS

X such that fK (xi ) = y. Thus the analog of the extension property is the propertythat the map X(K) → Y (K) is surjective and has finite fibres. Let B is generatedover A by one element b satisfying an algebraic equation

a0 xn + a1 xn−1 + . . . + an = 0

with coefficients in A. Assume the ideal I = (a0 , . . . , an−1 ) is proper but an is

invertible in A. Let m be a maximal ideal in A containing I. Let K be an algebraicallyclosed field containing the residue field A/m. Consider the K-point of Y correspondingto the homomorphism y : A → A/m → K. Since B ∼ = A[x]/(a0 xn + a1 xn−1 + . . . +an ), any homomorphism extending y must send an to zero but this is impossible sincean is invertible. Other bad thing may happen if an ∈ I. Then we obtain infinitelymany extensions of y, they are defined by sending x to any element in K. It turns outthat requiring that a0 is invertible will guarantee that X(K) → Y (K) is surjectivewith finite fibres. We start with reviewing some facts from commutative algebra.

Definition 10.1. A commutative algebra B over a commutative ring A is said to be

integral over A if every element x ∈ B is integral over A (i.e. satisfies an equationxn + a1 xn−1 + . . . + an = 0 with ai ∈ A).

Lemma 10.1. Assume that B is a finitely generated A-algebra. Then B is integral

(i.e., for any b ∈ B there exists F ∈ A[Z1 , . . . , Zn ] such that b = F (x1 , . . . , xn )). n(i)Since each xi is integral over A, there exists some integer n(i) such that xi can bewritten as a linear combination of lower powers of xi with coefficients in A. Henceevery power of xi can be expressed as a linear combination of powers of xi of degreeless than n(i). Thus there exists a number N > 0 such that every b ∈ B can bewritten as a polynomial in x1 , . . . , xn of degree < N . This shows that a finite set ofmonomials in x1 , . . . , xn generate B as an A-module. Conversely, assume that B is a finitely generated A-module. Then every b ∈ Bcan be written as a linear combination b = a1 b1 + . . . + ar br , where b1 , . . . , br is afixed set of elements in B and ai ∈ A. Multiplying the both sides by bi and expressingeach product bi bj as a linear combination of bi ’s we get X bbi = aij bi , aij ∈ A. (10.1) j

Proof. We know that this is true for an affine set X (see Lecture 8). Let X beany quasi-projective algebraic k-set. Obviously, for any open affine set U we haveD(φ|U ) = U ∩ D(φ). This shows that φ|U ∩ D(φ) is invertible, and by taking anaffine open cover of D(φ), we conclude that φ|D(φ) is invertible. By the universalproperty of localization, this defines a homomorphism α : O(X)φ → O(D(φ)). The re-striction homomorphism O(X) → O(U ) induces the homomorphism αU : O(U )φ|U →O(D(φ) ∩ U ). By taking an affine open cover of X = ∪i Ui , we obtain that all αUiare isomorphisms. Since every element of O(X) is uniquely determined by its restric-tions to each Ui , and any element of O(D(φ)) is determined by its restriction to eachD(φ) ∩ Ui , we obtain that α is an isomorphism.

Lemma 10.6. Let X and Y be two quasi-projective algebraic k-sets. Assume that Yis affine. Then the natural map

Mapreg (X, Y ) → Homk−alg (O(Y ), O(X)), f → f ∗,

is bijective.

Proof. We know this already if X and Y are both affine. Let U be an affine opensubset of X. By restriction of maps (resp. functions), we obtain a commutativediagram: Mapreg (X, Y ) / Homk−alg (O(Y ), O(X))

Mapreg (U, Y ) / Homk−alg (O(Y ), O(U )).

Here the bottom horizontal arrow is a bijection. Thus we can invert the upper horizon-tal arrow as follows. Pick up an open affine cover {Ui }i∈I of X. Take a homomorphismφ : O(Y ) → O(X), its image in Homk−alg (O(Y ), O(Ui )) is the composition with therestriction map O(X) → O(Ui ). It defines a regular map Ui → Y . Since a regularmap is defined on its open cover, we can reconstruct a “global” map X → Y . It iseasy to see that this is the needed inverse.

Lemma 10.7. Let X be a quasi-projective algebraic k-set. Then X is affine if and

only if O(X) is a finitely generated k-algebra which contains a finite set of elementsφi which generate the unit ideal and such that each D(φi ) is affine. 89

(ii) for any y ∈ Y , the fibre f −1 (y) is a finite set.

Proof. Clearly, we may assume that X and Y are affine, B = O(X) is integralover A = O(Y ) and φ = f ∗ is injective. A point y ∈ Y defines a homomorphismevy : A → K whose kernel is a prime ideal p. A point x ∈ f −1 (y) correspondsto a homomorphism evx : B → K of k-algebras such that its composition withφ is equal to evy . By Lemma 10.3 (vi), there exists a prime ideal P in B suchthat φ−1 (P) = p. Let Q(B/P) be the field of fractions of the quotient ring B/Pand Q(A/p) be the field of fractions of the ring A/p. Since B is integral over A,the homomorphism φ defines an algebraic extension Q(B/P)/Q(A/p) (Lemma 10.3(iv)). Since K is algebraically closed, there exists a homomorphism Q(B/P) → Kwhich extends the natural homomorphism Q(A/p) → K defined by the injectivehomomorphism A/p → K induced by evy . The composition of the restriction of thehomomorphism Q(B/P) → K to B/P and the factor map B → B/P defines a pointx ∈ f −1 (y). This proves the surjectivity of f . Note that the field extension Q(B/P)/Q(A/p) is finite (since it is algebraic andQ(B/P) is a finitely generated algebra over Q(A/p). It is known from the theory 91

of field extensions that the number of homomorphisms Q(B/P) → K extending the

So it suffices to show that the number of prime ideals P ⊂ B such that φ−1 (P) = pis finite. It follows from Lemma 10.3 (iii) that the set of such prime ideals is equalto the set of irreducible components of the closed subset of X defined by the properideal pB. We know that the number of irreducible components of an affine k-set isfinite. This proves the second assertion.

Proof. Assume first that X is projective. Let X be a closed subset of some Prk (K).If X = Prk (K), we take for f the identity map. So we may assume that X is aproper closed subset. Since k is infinite, we can find a point x ∈ Prk (k) \ X. Letpx : X → Pr−1 k (K) be the linear projection from the point x. We know from theprevious examples that px : X → px (X) is a finite k-map. If px (X) = Pr−1 k (K), weare done. Otherwise, we take a point outside px (X) and project from it. Finally, weobtain a finite map (composition of finite maps) X → Pnk (K) for some n. Assume that X is affine. Then, we replace X by an isomorphic set lying as a closedsubset of Prk (K)0 of some Prk (K). Let X̄ be the closure of X in Prk (K). Projectingfrom a point x ∈ Prk (K) \ (X̄ ∪ Prk (K)0 ), we define a finite map X̄ → Pr−1 k (K). Sinceone of the equations defining x can be taken to be T0 = 0, the image of Prk (K)0 iscontained in Pr−1 r−1 ∼ r−1 k (K)0 . Thus the image of X is contained in Pk (K)0 = Ak (K).Continuing as in the projective case, we prove the theorem.

The next corollary is called the Noether Normalization theorem. Together with thetwo Hilbert’s theorems (Basis and Nullstellensatz) these three theorems were knownas “the three whales of algebraic geometry.”

Corollary 10.11. Let A be a finitely generated algebra over a field k. Then A is

(b) X = Y = A2 , f is defined by the formula (x, y) → (xy, y).

2. Let f : X → Y be a finite map. Show that the image of any closed subset of X isclosed in Y .3. Let f : X → Y and g : X 0 → Y 0 be two finite regular maps. Prove that theCartesian product map f × g : X × X 0 → Y × Y 0 is a finite regular map.4. Give an example of a surjective regular map with finite fibres which is not finite.5. Let A be an integral domain, Q be its field of fractions. The integral closure Ā ofA in Q is called the normalization of A. A normal ring is a ring A such that A = Ā.

(i) dim X = 0 if X is a non-empty Hausdorff space;

(iii) dim X ≥ dim Y if Y ⊂ X, the strict inequality takes place if none of the irreducible components of the closure of Y is an irreducible component of X;

9394 LECTURE 11. DIMENSION

(iv) if X is covered by a family of open subsets Ui , then dim X is equal to supi Ui .

Proof. (i) In a non-empty Hausdorff space a point is the only closed irreducible subset. (ii) Let Z0 ⊃ Z1 ⊃ . . . ⊃ Zr be a strictly decreasing chain of irreducible closedsubsets of X. Then Z0 = ∪i∈I (Z0 ∩ Xi ) is the union of closed subsets Z0 ∩ Xi . SinceZ0 is irreducible, Z0 ∩ Xi = Z0 for some Xi , i.e., Z0 ⊂ Xi . Thus the above chain isa chain of irreducible closed subsets in Xi and r ≤ dim Xi . (iii) Let Z0 ⊃ Z1 ⊃ . . . ⊃ Zr be a strictly decreasing chain of irreducible closedsubsets of Y , then the chain of the closures Z̄i of Zi in X of these sets is a strictlydecreasing chain of irreducible closed subsets of X. As we saw in the proof of (ii) allZ̄i are contained in some irreducible component Xi of X. If this component is a notan irreducible component of the closure of Y , then Xi ⊃ Z̄0 and we can add it to thechain to obtain that dim X > dim Y .

(iv) Left to the reader.

Proposition 11.3. An algebraic k-set X is of dimension 0 if and only if it is a finite

set.

Proof. By Proposition 11.2 (ii) we may assume that X is irreducible. Suppose

dim X = 0. Take a point x ∈ X and consider its closure Z in the Zariski k-topology. Itis an irreducible closed subset which does not contain proper closed subsets (if it does,we find a proper closed irreducible subset of Z). Since dim X = 0, we get Z = X. Wewant to show that X is finite. By taking an affine open cover, we may assume that X isaffine. Now O(X) is isomorphic to a quotient of polynomial algebra k[Z1 , . . . , Zn ]/I.Since X does not contain proper closed subsets I must be a maximal ideal. As we sawin the proof of the Nullstellensatz this implies that O(X) is a finite field extension of k.Every point of X is defined by a homomorphism O(X) → K. Since K is algebraicallyclosed there is only a finite number of homomorphisms O(X) → K. Thus X is afinite set (of cardinality equal to the separable degree of the extension O(X)/k). Conversely, if X is a finite irreducible set, then X is a finite union of the closuresof its points. By irreducibility it is equal to the closure of any of its points. Clearly itdoes not contain proper closed subsets, hence dim X = 0.

Definition 11.2. For every commutative ring A its Krull dimension is defined by

= alg.dimk Q(A) = alg.dimk A.

So we see that for irreducible affine algebraic sets the following equalities hold:

dim X = dim O(X) = alg.dimk O(X) = alg.dimk R(X) = n

where n is defined by the existence of a finite map X → Ank (K).

Note that, since algebraic dimension does not change under algebraic extensions,we obtain

Corollary 11.9. Let X be an affine algebraic k-set and let X 0 be the same set con-sidered as an algebraic k 0 -set for some algebraic extension k 0 of k. Then

dim X = dim X 0 .

To extend the previous results to arbitrary algebraic sets X, we will show that forevery dense open affine subset U ⊂ X

dim U = dim X. (11.1)

If X is an affine irreducible set, and U = D(φ) for some φ ∈ O(X), then it is easyto see. We have

dim D(φ) = dim O(U )[ φ1 ] = alg.dimk Q(O(U )[ φ1 ]) (11.2)

= alg.dimk Q(O(U )) = dim U.

It follows from this equality that any two affine sets have the same dimension(because they contains a common subset of the form D(φ)). To prove equality (11.1) in the general case we need the following.

F (T )0 = T n + (a1 x/y)T n−1 + . . . + (an xn /y n ) = 0

with coefficients in the field Q(A). If u satisfies an equation of smaller degree overQ(A), after plugging in u = xz/y, we find that z satisfies an equation of degreesmaller than n. This is impossible by the choice of F (T ). Thus F (T )0 is a minimalpolynomial of u. Since u is integral over A, the coefficients of F (T )0 belong to A.Therefore, y i |ai xi , and, since x and y are coprime, y i |ai . This implies that un +xt = 0for some t ∈ A, and therefore x|un .

Theorem 11.13. Let X be an algebraic set and U be a dense open subset of X.

Then dim X = dim U.

Proof. Obviously, we may assume that X is irreducible and U is its open subset. Firstlet us show that all affine open subsets of X have the same dimension. For this it isenough to show that dim U = dim V if V ⊂ U are affine open subsets. Indeed, weknow that for every pair U and U 0 of open affine subsets of X we can find an affine non-empty subset W ⊂ U ∩ U 0 . Then the above will prove that dim W = dim U, dim W =dim U 0 . Assume U is affine, we can find an open non-empty subset D(φ) ⊂ V ⊂ U ,where φ ∈ O(U ) \ O(U )∗ . Applying (11.2), we get dim U = dim D(φ). This shows100 LECTURE 11. DIMENSION

that all open non-empty affine subsets of X have the same dimension. Let Z0 ⊃Z1 ⊃ . . . ⊃ Zn be a maximal decreasing chain of closed irreducible subsets of X, i.e.,n = dim X. Take x ∈ Zn and let U be any open affine neighborhood of x. Then

Z0 ∩ U ⊃ Z1 ∩ U ⊃ . . . ⊃ Zn ∩ U 6= ∅

is a decreasing chain of closed irreducible subsets of U (note that Zi ∩ U 6= Zj ∩ U for

dim Y = dim Y ∩ U = dim U − 1 = dim X − 1.

Y = X ∩ V ((F1 , . . . , Fr )) = X ∩ V (F1 ) ∩ . . . ∩ V (Fr )

101

be the set of its common zeroes and Z be an irreducible component of this set. Then,either Z is empty, or dim Z ≥ dim X − r.The equality takes place if and only if for every i = 1, . . . , r the polynomial Fi doesnot vanish identically on any irreducible component of X ∩ V (F1 ) ∩ . . . ∩ V (Fi−1 ).

dim X × Y = dim X + dim Y.

Proof. Consider the projection X × Y → Y and apply the Theorem.

dim Z ≥ dim X + dim Y − n.

Proof. Replacing X and Y by its open affine subsets, we may assume that X and Yare closed subsets of An (K). Let ∆ : An (K) → An (K)×An (K) be the diagonal map.Then ∆ maps X ∩ Y isomorphically onto (X × Y ) ∩ ∆An (K) , where ∆An (K) is thediagonal of An (K). However, ∆An (K) is the set of common zeroes of n polynomialsZi − Zi0 where Z1 , . . . , Zn are coordinates in the first factor and Z10 , . . . , Zn0 are thesame for the second factor. Thus we may apply Theorem 11.10 n times to obtain

dim Z ≥ dim X × Y − n.

It remains to apply the previous corollary.

We define the codimension codim Y (or codim (Y, X) to be precise) of a subspace

Y of a topological space X as dim X − dim Y . The previous theorem can be statedin these terms as

codim (X ∩ Y, Pn (K)) ≤ codim (X, Pn (K)) + codim (Y, Pn (K)).

In this way it can be stated for the intersection of any number of subsets.

Problems.1. Give an example of

(a) a topological space X and its dense open subset U such that dim U < dim X;

(c) a Noetherian topological space of infinite dimension.

2. Prove that every closed irreducible subset of Pn (K) or An (K) of codimension 1 is

the set of zeroes of one irreducible polynomial.3. Let us identify the space K nm with the space of matrices of size m × n with entriesin K. Let X 0 be the subset of matrices of rank ≤ m − 1 where m ≤ n. Show thatthe image of X 0 \ {0} in the projective space Pnm−1 (K) is an irreducible projectiveset of codimension n − m + 1.4. Show that for every irreducible closed subset Z of an irreducible algebraic set Xthere exists a chain of n = dim X + 1 strictly decreasing closed irreducible subsets104 LECTURE 11. DIMENSION

containing Z as its member. Define codimension of an irreducible closed subset Z of

Prove that dim Y + codim (Y, X) = dim X. In particular, our definition agrees withthe one given at the end of this lecture.5. A subset V of a topological space X is called constructible if it is equal to a disjointunion of finitely many locally closed subsets. Using the proof of Theorem 11.19 showthat the image f (V ) of a constructible subset V ⊂ X under a regular map f : X → Yof quasi-projective sets contains a non-empty open subset of its closure in Y . Usingthis show that f (V ) is constructible (Chevalley’s theorem).6*. Let X be an irreducible projective curve in Pn (K), where k = K, and E =V (a0 T0 + . . . + an Tn ) be a linear hyperplane. Show that E intersects X at the samenumber of distinct points if the coefficients (a0 , . . . , an ) belong to a certain Zariskiopen subset of the space of the coefficients. This number is called the degree of X.7*. Show that the degree of the Veronese curve vr (P1 (K)) ⊂ Pn (K) is equal to r.8*. Generalize Bezout’s Theorem by proving that the set of solution of n homogeneousequations of degree d1 , . . . , dn is either infinite or consists of d1 · · · dn points takenwith appropriate multiplicities.Lecture 12

Lines on hypersurfaces

In this lecture we shall give an application of the theory of dimension. Consider thefollowing problem. Let X = V (F ) be a projective hypersurface of degree d = degFin Pn (K). Does it contain a linear subspace of given dimension, and if it does,how many? Consider the simplest case when d = 2 (the case d = 1 is obviouslytrivial). Then F is a quadratic form in n + 1 variables. Let us assume for simplicitythat char(K) 6= 2. Then a linear m-dimensional subspace of dimension in V (F )corresponds to a vector subspace L of dimension m + 1 in K n+1 contained in the setof zeroes of F in K n+1 . This is an isotropic subspace of the quadratic form F . Fromthe theory of quadratic forms we know that each isotropic subspace is contained ina maximal isotropic subspace of dimension n + 1 − r + [r/2], where r is the rank ofF . Thus V (F ) contains linear subspaces of dimension ≤ n − r + [r/2] but does notcontain linear subspaces of larger dimension. For example, if n = 3, and r = 4, X isisomorphic to V (T0 T1 − T2 T3 ). For every λ, µ ∈ K, we have a line L(λ, µ) given bythe equations λT0 + µT2 = 0, µT1 + λT3 = 0,or a line M (λ, µ) given by the equation

see that I(r, d, n) is given by bi-homogeneous polynomials in the coefficients of F andin the Plücker coordinates of E. This proves that I(r, d, n) is a closed subset of theproduct H × G. Now consider the projection p : I(r, d, n) → G. For each E ∈ G, thefibre p−1 (E) consists of all hypersurfaces V (F ) containing E. Choose a coordinatesystem such that E is given by the equations Tr+1 = . . . = Tn = 0. Then E ⊂ V (F )if and only if each monomial entering into F with non-zero coefficient contains somepositive power of Ti with i ≥ r + 1. In other words F is defined by vanishing of allcoefficients at the monomials of degree d in the variables T0 , . . . , Tr . This gives r+d rlinear conditions on the coefficients of F , hence dim p−1 (E) = n+d r+d

d − 1 − r .Let us assume that I(r, d, n) is irreducible. By the Theorem on dimension of fibres,

d = 4, so we expect that every cubic surface has a line. Aswe saw in the previous remark it needs to be proven.

Theorem 12.10. (i) Every cubic surface X contains a line.

(ii) There exists an open subset U ⊂ Hyp(3; 3)(K) such that any X ∈ U contains exactly 27 lines.

Proof. (i) In the notation of the proof of Theorem 12.8, it suffices to show that theprojection map q : I(1, 3, 3) → Hyp(3; 3)(K) is surjective. Suppose the image of q isa proper closed subset Y of Hyp(3; 3)(K). Then dim Y < dim Hyp(3; 3)(K) = 19and dim I(1, 3, 3) = 19. By the theorem on dimension of fibres, we obtain that allfibres of q are of dimension at least one. In particular, every cubic surface containinga line contains infinitely many of them. But let us consider the surface X given bythe equation T1 T2 T3 − T03 = 0.Suppose a line ` lies on X. Let (a0 , a1 , a2 , a3 ) ∈ `. If a0 6= 0, then ai 6= 0, i 6= 0.On the other hand, every line hits the planes Ti = 0. This shows that ` is containedin the plane T0 = 0. But there are only three lines on X contained in this plane:Ti = T0 = 0, i = 1, 2 and 3. Therefore X contains only 3 lines. This proves the firstassertion. (ii) We already know that every cubic surface X = V (F ) has at least one line.Pick up such a line `. Without loss of generality we may assume that it is given bythe equation: T2 = T3 = 0. 113

As we saw in the proof of Lemma 12.6:

F = T2 Q0 (T0 , T1 , T2 , T3 ) + T3 Q1 (T0 , T1 , T2 , T3 ) = 0,

where Q0 and Q1 are quadratic homogeneous polynomials. Each plane π containing

Then C(λ, µ) is given by the equation:

+(µ(µa02 +λb02 )+λ(µa03 +λb03 ))t0 t2 +(µ2 a12 +λµb12 )+((µλa13 +λ2 b13 ))t1 t2 = 0. Now, let us start vary the parameters λ and µ and see how many reducible conicsC(λ, µ) we obtain. The conic C(λ, µ) is reducible if and only if the quadratic formdefining it is degenerate. The condition for the latter is the vanishing of the discrimi-nant D of the quadratic form C(λ, µ). Observe that D is a homogeneous polynomialof degree 5 in λ, µ. Thus there exists a Zariski open subset of Hyp(3; 3) for which thisdeterminant has 5 distinct roots (λi , µi ). Each such solution defines a plane πi whichcut out on X the line ` and a reducible conic. The latter is the union of two lines ora double line. Again, for some open subset of Hyp(3; 3) we expect that the double114 LECTURE 12. LINES ON HYPERSURFACES

line case does not occur. Thus we found 11 lines on X: the line ` and 5 pairs of lines`i , `0i lying each lying in the plane πi . Pick up some plane, say π1 . We have 3 lines`, `0 , and `00 in π1 . Replacing ` by `0 , and then by `00 , and repeating the construction,we obtain 4 planes through `, `0 and 4 planes through `00 , each containing a pair oflines. Altogether we found 3 + 8 + 8 + 8 = 27 lines on X. To see that all lines areaccounted for, we observe that any line intersecting either `, or `0 , or `00 lies in one ofthe planes we have considered before. So it has been accounted for. Now let L be anyline. We find a plane π through L that contains three lines L, L0 and L00 on X. Thisplane intersects the lines `, `0 , and `00 at some points p, p0 and p00 respectively. Wemay assume that these points are distinct. Otherwise we find three no-coplanar linesin X passing through one point. As we shall see later this implies that X is singularat this point. Since neither L0 nor L00 can pass through two of these points, one ofthese points lie on L. Hence L is coplanar with one of the lines `, `0 , `00 . Therefore Lhas been accounted for.

Remark 12.11. Using more techniques one can show that every “nonsingular” (in thesense of the next lectures) cubic surface contains exactly 27 lines. Let us define thegraph whose vertices are the lines and two vertices are joined by an edge if the linesintersect. This graph is independent on the choice of a nonsingular cubic surface andits group of symmetries is isomorphic to the group W (E6 ) of order 51840 (the Weylgroup of the root system of a simple Lie algebra of type E6 ).

Problems.1. Show that the set πx of lines in P3 (K) passing through a point x ∈ P3 (K) is aclosed subset of G(2, 4) isomorphic to P2 (K). Also show that the set πP of lines inP3 (K) contained in a plane P ⊂ P3 (K) is a closed subset of G(2, 4) isomorphic toP2 (K).2. Prove that the subset of quartic surfaces in Hyp(4; 3) which contain a line is anirreducible closed subset of Hyp(4; 3) of codimension 1.3. Prove that every hypersurface of degree d ≤ 5 in P4 (K) contains a line, and, ifd ≤ 4, then it contains infinitely many lines.4. Let X be a general cubic hypersurface in P4 (K) (general means that X belongs toan open subset U of Hyp(3; 4)). Show that there exists an open subset V ⊂ X suchthat any x ∈ V lies on exactly six lines contained in X.5*. Let 1 ≤ m1 < m2 < . . . < mr ≤ n + 1, a flag in K n+1 of type (m1 , . . . , mr ) is achain of linear subspaces L1 ⊂ . . . ⊂ Lr with dim Li = mi . (a) Show that the set of flags is a closed subset in the product of the Grassmannians G(m1 , n + 1) × . . . × G(mr , n + 1). This projective algebraic set is called the flag variety of type (m1 , . . . , mr ) and is denoted by Fk (m1 , . . . , mr ; n + 1). 115

(b) Find the dimension of Fk (m1 , . . . , mr ; n + 1).

6. By analyzing the proof of Theorem 13.16 show the following:

(a) The set of 27 lines on a cubic surface X contains 45 triples of lines which lie in a plane (called a tritangent plane).

7*. Prove that

(b) Using (a) show that V (F ) can be given by the equation

  L1 0 M1 det M2 L2 0  = 0, 0 M3 L3

where Li , Mi are linear forms.

(c) Show that the map T : V (F ) → P2 which assigns to a point x ∈ V (F ) the set of solutions of the equation (t0 , t1 , t2 ) · A = 0 is a birational map. Here A is the matrix of linear from from (b).

(d) Find an explicit formulas for the inverse birational map T −1 .

8. Using Problem 5 (b),(c) show that the group W of symmetries of 27 lines consistsof 51840 elements.9*. Let C be a twisted cubic in P3 (the image of P1 under a Veronese map givenby monomials of degree 3). For any two distinct point x, y ∈ C consider the linelx,y joining these points. Show that the set of such lines is a locally closed subset ofG(2, 4). Find the equations defining its closure.10*. Let k = k0 (t), where k0 is an algebraically closed field and F (T0 , . . . , Tn ) ∈k[T1 , . . . , Tn ] be a homogeneous polynomial of degree d < n. Show that V (F )(k) 6= ∅(Tsen’s Theorem).116 LECTURE 12. LINES ON HYPERSURFACESLecture 13

Tangent space

The notion of the tangent space is familiar from analytic geometry. For example, letF (x, y) = 0 be a curve in R2 and let a = (x0 , y0 ) be a point lying on this surface.The tangent line of X at the point a is defined by the equation: ∂F ∂F (a)(x − x0 ) + (a)(y − y0 ) = 0. ∂x ∂yIt is defined only if at least one partial derivative of F at a is not equal to zero. Inthis case the point is called nonsingular. Otherwise it is said to be singular. Another notion of the tangent space is familiar from the theory of differentiablemanifolds. Let X be a differentiable manifold and a be its point. By definition, atangent vector ta of X at a is a derivation (or differentiation) of the ring O(X) ofdifferentiable functions on X, that is, a R-linear map δ : O(X) → R such that δ(f g) = f (a)δ(g) + g(a)δ(f ) for any f, g ∈ O(X).It is defined by derivation of a function f along ta given by the formula n X ∂f < f, ta >= (a)ti ∂xi i=1

where (t1 , . . . , tn ) are the coordinates of ta and (x1 , . . . , xn ) are the local coordinatesof X at the point a. In this lecture we introduce and study the notion of a tangent space and a non-singular point for arbitrary algebraic sets or varieties. For every k-algebra K let K[ε] = K[t]/(t2 ) be the K-algebra of dual numbers. Ifε is taken to be t mod(t2 ), then K[ε] consists of linear combinations a + bε, a, b ∈ K,which are added coordinate wise and multiplied by the rule (a + bε)(a0 + b0 ε) = aa0 + (ab0 + a0 b)ε.

Definition 13.1. Let X be an affine or a projective algebraic variety over a field k andx ∈ X(K) be its K-point. A tangent vector tx of F at x is a K[ε]-point tx ∈ X(K[ε])such that X(α1 )(tx ) = x. The set of tangent vectors of X at x is denoted by T (X)xand is called the tangent space of F at x.

Example 13.1. Assume X is an affine algebraic variety given by a system of equations

F1 (Z1 , . . . , Zn ) = 0, . . . , Fr (Z1 , . . . , Zn ) = 0.

A point x ∈ X(K) is a solution (a1 , . . . , an ) ∈ K n of this system. A tangent vector

Thus the set of tangent vectors T (X)x is bijective to the submodule of K n whichconsists of solutions of a homogeneous system of linear equations. In particular, wehave introduced the structure of a K-module on T (X)x . 119

Let Lx be the line in K n+1 corresponding to x. We see that a tangent vector tx

T (Pnk )x → Homk (Lx , K n+1 /Lx ).

Since the right-hand side has a natural structure of a rank n free module over K, wecan transfer this structure to T (Pnk )x .

Example 13.3. Let X = GLn,k be the affine algebraic variety with GLn,k (K) =GL(n, K). A point of GLn,k (K[ε]) is a matrix A + εB, where A ∈ GL(n, K), B ∈Matn (K). If we take x ∈ X(K) to be the identity matrix In , we obtain that T (X)In can beidentified with Matn (K). Now, take X = SLn,k with X(K) = SL(n, k). Then

T (X)In = {In + εB ∈ GL(n, K[ε]) : det(In + εB) = 1} =

{In + εB ∈ GL(n, K[ε]) : Trace(B) = 0}.

Thus we can identify T (SLn,k )In with the vector space of matrices with entries in Kwith trace equal to zero. Now let us take X = On,k with On,k = {A ∈ Matn (K) : A · t A = In }. We get

T (X)In = {In + εB ∈ Mat(n, K[ε]) : (In + εB)(In + εt B) = In } =

{In + εB ∈ GL(n, K[ε]) : B + t B = 0}.

120 LECTURE 13. TANGENT SPACE

Thus we can identify T (On,k )In with the vector space of skew-symmetric matriceswith entries in K. Note that the choice of K depends on identification of In with aK-point. The tangent space of an algebraic group at the identity point has a structure of aLie algebra defined by the Lie bracket.

Remark 13.4. For any functor F from the category of k-algebras to the category ofsets one can define the tangent space of F at a “point” x ∈ F (K) as the set ofelements t of the set F (K[ε])(t) = x. Now, if we have a projective variety X given by a system of homogeneous equationsF1 = · · · = Fk = 0, we obtain that

where di is the degree of Fi .

As we saw the tangent space of an affine or a projective variety has a structure ofa linear space. However, it is not clear that this structure is independent of a choice ofthe system of equations defining X. To overcome this difficulty, we shall give another,more invariant, definition of T (X)x . Let A be a commutative k-algebra and let M be A-module. A M -derivation ofA is a linear map of the corresponding k-linear spaces δ : A → M such that for alla, b ∈ A δ(ab) = aδ(b) + bδ(a). 121

The set of M -derivations is denoted by Derk (A, M ). It has a natural structure of a

A-module via (aδ)(b) = aδ(b) for all a, b ∈ A.We will be interested in a special case of this definition.

Lemma 13.5. If f : A → B is a homomorphism of k-algebras, and δ : B → M is a

M -derivation of B, then the composition δ ◦ f : A → B → M is a M[f ] -derivationof A, where M[f ] is the A-module obtained from M by the operation of restriction ofscalars (i.e., a · m = f (a)m for any a ∈ A, m ∈ M ).

Proof. Trivial verification of the definition.

Let us apply this to our situation. Note that the k-linear map:

α2 : K[ε] → K, a + bε → b

is a K-derivation of K[ε] considered as a K-algebra. Here K is considered as a K[ε]-

defined in Lemma 13.5, where f is the homomorphism f ∗ : O(Y ) → O(X). This is

d(g ◦ f )x = (dg)y ◦ (df )x .

Proof. Immediately follows from the definition of a morphism.

Now we can define the tangent space for any quasi-projective algebraic set V ⊂Pn (K). Here K, as usual, is a fixed algebraically closed field containing k. First, weassume that V is affine. Choose an affine algebraic K-variety X such that I(X) isradical and X(K) = V . Then we define the the tangent space T (V )x of V at x bysetting T (V )x = T (X)x .By Lemma 13.7, for every x ∈ V we have an isomorphism of K-linear spaces.

T (X)x ∼ = Derk (O(X), K).Since an isomorphism of affine varieties is defined by an isomorphism of their coordinatealgebras, we see that this definition is independent (up to isomorphism of linear spaces)of a choice of equations defining X.Lemma 13.9. Let A be a commutative K-algebra, M an A-module, and S a multi-plicatively close subset of A. There is an isomorphism of AS -modules

Derk (A, M )S ∼ = Derk (AS , MS )Proof. Let δ : A → M be a derivation of A. We assign to it the derivation of ASdefined by the familiar rule: a δ(a)s − δ(s)a δ = . s s2This definition does not depend on the choice of a representative of the fraction as . Infact, assume s00 (s0 a − sa0 ) = 0. Then

Multiplying both sides by s00 , we obtain

s002 [δ(s0 a) − δ(sa0 )] = 0. (13.4)

Let us show that this implies that

This will proves our assertion. The previous identity is equivalent to the following one

s002 [s2 s0 δ(a0 ) − s02 sδ(a)] = s002 [s2 a0 δ(s0 ) − s02 aδ(s)],

or s0 ss002 [sδ(a0 ) − s0 δ(a)] = ss0 s002 [aδ(s0 ) − a0 δ(s)].Now this follows from equality (13.4) after we multiply it by ss0 . So we have defined a homomorphism of A-modules Derk (A, M ) → Derk (AS , MS ).It induces a map of AS -modules Derk (A, M )S → Derk (AS , MS ). The inverse of thismap is defined by using Lemma 13.5 applied to the homomorphism A → AS .

Let us apply the previous lemma to our situation. Let X be an affine k-variety,x ∈ X(K), p = Ker(evx ). Assume that K is a field. Then the ideal p is primesince O(X)/p is isomorphic to a subring of K. Consider K as a module over O(X)by means of the homomorphism evx . Let S = O(X) \ p. Then KS = K since theimage of S under evx does not contain 0. It is easy to see that the linear K-spacesDer(O(X), K)p and Der(O(X), K) are isomorphic (the map ∂s 7→ evx (s)−1 ∂ is theisomorphism). Applying Lemma 13.9, we obtain an isomorphism of vector K-spaces

T (X)x = Derk (O(X), K)x ∼

= Derk (O(X)p , K). (13.5)

The previous isomorphism suggests a definition of the tangent space of any quasi-projective algebraic k-set X.

Definition 13.2. The local ring of X at x ∈ X is the factor set

[ OX,x = O(U )/R x∈U

where U runs through the set of all open affine neighborhoods of x and the equivalencerelation R is defined as follows:

Let f ∈ O(U ), g ∈ O(V ), then

We shall call the equivalence class of f ∈ O(U ), the germ of f at x. The structure ofa ring in OX,x is induced by the ring structure of any O(U ). We take two elements ofOX.x , represent them by regular functions on a common open affine subset, multiplyor add them, and take the germ of the result. Let mX,x be the ideal of germs offunctions f ∈ O(U ) which vanish at x. It follows from the definition that, for any open affine neighborhood U of x, thenatural map O(U ) → OX,x , φ 7→ φx defines an isomorphism

OU,x ∼ = OX,x .

Lemma 13.10. (i) mX,x is the unique maximal ideal of OX,x .

(ii) If X is affine and irreducible, the canonical homomorphism O(X) → OX,x

Proof. (i) It suffices to show that every element α ∈ OX,x \ mX,x is invertible. Letα = fx , where f is regular on a some open affine set U containing x. Since f (x) 6= 0, xis contained in the open principal affine subset V = D(f ) of U . Hence the restrictiong of f to V is invertible in O(V ). The germ gx = fx is now invertible. (ii) For any φ ∈ O(X) \ px its germ in OX,x is invertible. By the universalproperty of localizations, this defines a homomorphism O(X)px → OX,x . An elementof the kernel of this homomorphism is a function whose restriction to some openneighborhood of x is identically zero. Since X is irreducible, this implies that thefunction is zero. Let fx ∈ OX,x be the germ of a function f ∈ O(U ), where U is anopen affine neighborhood of x. Replacing U by a principal open subset D(φ) ⊂ U ,we may assume that U = D(φ) and f = F/φn , where F, φ ∈ O(X). Since φ(x) 6= 0,we get that φ does not belong to px , and hence f ∈ O(X)px and its germ at x equalsfx . This proves the surjectivity. (iii) This follows easily from the definition of the localization ring Ap for any ringA and a prime ideal p. The homomorphism A → Ap , a 7→ a1 defines a homomorphismA/p → Ap /pAp . The target space is a field. By the universal property of fields offractions, we get a homomorphism of fields g : Q(A/p) → Ap /pAp . Let as + pAp ∈Ap /pAp . Then it is the image of the fraction a+p s+p ∈ Q(A/p). This shows that g isbijective.

In the case when x ∈ X(k), choosing an open affine neighborhood of x and applyingLemma 13.7, we obtain

T (X)x = Homk (mX,x /m2X,x , k). (13.7)

For any rational point x ∈ X(k), the right-hand side of (13.7) is called the Zariskitangent space of X at x. Let f : X → Y be a regular map of algebraic k-sets, x ∈ X and y = f (x). LetV be an open affine neighborhood of y and U be an open affine neighborhood of xcontained in f −1 (V ). The restriction of f to U defines a regular map f : U → V . Forany φ ∈ O(V ), the composition with f defines a regular function f ∗ (φ) on U . Letf ∗ (φ)x ∈ OU,x be its germ at x. The homomorphism f ∗ : O(V ) → OU,x extends to ahomomorphism f ∗ : OV,y → OU,x of the local rings. It is clear that f ∗ (mV,y ) ⊂ mU,x .Composing this homomorphism with the isomorphisms OX,x ∼ = OU,x and OY,y ∼ = OV,ywe get a homomorphism of local rings ∗ fx,y : OY,y → OX,x . (13.8)

Applying Lemma 13.5, we get a K-linear map

TX,x = Derk (OX,x , K) → TY,y = Derk (OY,y , K),

which we call the differential of f at the point x and denote by dfx .

Let X ⊂ Pnk (K) be an quasi-projective algebraic k-set and T (X)x be the tangentspace at its point x ∈ X(K). It is a vector space over K of finite dimension. In fact,it is a subspace of T (Pn (K))x ∼ = K n and hence

dimK T (X)x ≤ n. (13.9)

If x is contained in an affine open subset U which is isomorphic to a closed subset of

some An (K), then T (X)x is a subspace of T (An (K))x ∼ = K n and

dimK T (X)x ≤ n.

It follows from (13.9) that X is not isomorphic to a quasi-projective subset of Pnk (K)for any n < dimK T (X)x .

Let us now show that dimK T (X)x ≥ dim X for any irreducible algebraic set Xand the equality takes place for almost all points x (i.e., for all points belonging to aZariski open subset of X). For this, we may obviously assume that X is affine. Let V = X(K) for some affine variety defined by a radical ideal in k[Z1 , . . . , Zn ].The set T (V ) = X(K[ε]) is a subset of K[ε]n which can be thought as the vectorspace K 2n . It is easy to see that T (X) is a closed algebraic subset of K 2n and themap p = X(φ) : T (X) → X, is a regular map (check it !). Note that the fibre p−1 (x)is equal to the tangent space T (X)x . Applying the theorem about the dimension offibres of a regular map, we obtain

Proposition 13.12. There exists a number d such that

dimK T (X)x ≥ d

and the equality takes place for all points x belonging to an open subset of X.

Theorem 13.14. Let X be an irreducible algebraic set and d = min{T (X)x }. Then

d = dim X.128 LECTURE 13. TANGENT SPACE

Proof. Obviously, it suffices to find an open subset U of X where dimK T (X)x =

dim X for all x ∈ U . Replacing X by an open affine set, we may assume that X isisomorphic to an open subset of a hypersurface V (F ) ⊂ An (K) for some irreduciblepolynomial F (Theorem 4.10 of Lecture 4). This shows that we may assume thatX = V (F ). For any x ∈ X, the tangent space T (X)x is given by one equation ∂F ∂F (x)b1 + . . . + (x)bn = 0. ∂Z1 ∂ZnClearly, its dimension is equal to n − 1 = dim X unless all the coefficients are zeroes. ∂FThe set of common zeroes of the polynomials ∂Z i is a closed subset of An (K) contained ∂F ∂Fin each hypersurface V ( ∂Z i ). Obviously, ∂Zi 6∈ (F ) unless it is equal to zero (comparethe degrees). Now the assertion follows from Lemma 13.13. Obviously, the assertion of the previous theorem is not true for a reducible set. Tosee this it is sufficient to consider the union of two sets of different dimension. It iseasy to modify the statement to extend it to the case of reducible sets.Definition 13.3. The dimension of X at a point x is the maximum dimx X of thedimensions of irreducible components of X containing x.Corollary 13.15. Let X be an algebraic set and x ∈ X. Then dimK T (X)x ≥ dimx X.Proof. Let Y be an irreducible component of X containing x. Obviously, T (Y )x ⊂T (X)x . Hence dimx Y ≤ dimK T (Y )x ≤ dimK T (X)x .This proves the assertion.Definition 13.4. A point x of an algebraic set X is said to be nonsingular (or simple, orsmooth) if dimK T (X)x = dimx X. Otherwise, it is said to be singular. An algebraicset X is said to be nonsingular (or smooth) if all its points are nonsingular. OtherwiseX is said to be singular. We already know how to recognize whether a point is nonsingular.Theorem 13.16. (The Jacobian criterion of a nonsingular point). Assume that X isan affine algebraic k-set given by a system of equations F1 (Z) = . . . = Fr (Z) = 0 inAn (K). Then x ∈ X is nonsingular if and only if rk J(x) = n − dimx X, where  ∂F1 ∂F1  ∂Z1 (x) . . . ∂Zn (x)  ... ... ...  J(x) =  ... . ... ...  ∂Fr ∂Fr ∂Z1 (x) ... ∂Zn (x) 129

In this lecture we will give some other properties of nonsingular points. As usual wefix an algebraically closed field K containing k and consider quasi-projective algebraick-sets, i.e. locally closed subsets of projective spaces Pnk (K). Recall that a point x ∈ X is called nonsingular if dimK T (X)x = dimx X. Whenx ∈ X(k) is a rational point, we know that T (X)x ∼ = Homk (mX,x /m2X,x , k). Thus arational point is nonsingular if and only if

dimk mX,x /m2X,x = dimx X.

Let us see first that dimx X = dim OX,x . The number dim OX,x is denoted oftenby codimx X and is called the codimension of the point x in X. The reason is simple.If X is affine and px = Ker(evx ), then we have

dim Ap = sup{r : ∃ a chain p = p0 ) . . . ) pr of prime ideals in A}.

may assume that q0 is the maximal ideal m of A. Let pi be the pre-image of qi in Aunder the natural homomorphism A → Ap . Since p0 = p, we get a chain of primeideals p = p0 ⊃ . . . ⊃ pr . Conversely, any chain of such ideals in A generates anincreasing chain of prime ideals in Ap . It is easy to see that pi Aq = pi+1 Ap impliespi = pi+1 . This proves the assertion. In commutative algebra the dimension of Ap is called the height of the prime idealp.

(14.1)Recall that algdimk k(x) = dim O(X)/p. Any increasing chain of prime ideals inO(X)/p can be lifted to an increasing chain of prime ideals in O(X) beginning at p,and after adding a chain of prime ideals descending from p gives an increasing chain ofprime ideals in O(X). This shows that codimx X +algdimk k(x) ≤ dimx X. Togetherwith the inequality (14.1), we obtain the assertion.Corollary 14.3. Assume that k(x) is an algebraic extension of k. Then

In particular, for a rational point x we have

dimK T (X)x = embdimx X. (14.3)

Definition 14.2. A point x ∈ X is called regular if OX,x is a regular local ring, i.e.

embdimx X = codimx X.

Remark 14.5. We know that a rational point is regular if and only if it is nonsingular.In fact, any nonsingular point is regular (see next Remark) but the converse is nottrue. Here is an example. Let k be a field of characteristic 2 and a ∈ k which is nota square. Let X be defined in A2k (K) by the equation Z12 + Z23 + a = 0. Taking the √partial derivatives we see that ( a, 0) ∈ K 2 is a singular point. On the other hand,the ring OX,x is regular of dimension 1. In fact, the ideal p = Ker(evx ) is a maximalideal generated by the cosets of Z12 + a and Z2 . But the first coset is equal to thecoset of Z23 , hence p is a principal ideal generated by Z2 . Thus mX,x is generated byone element and OX,x is a regular ring of dimension 1.Remark 14.6. If x is not a rational point, equality (14.3) may not be true. For example,let k = C, K be the algebraic closure of the field k(t) and consider X = Ak (K). Apoint x = t defines the prime ideal p = {0} = Ker(evx ) (because t is not algebraicover k). The local ring OX,x is isomorphic to the field of fractions of k[Z1 ]. Hence itsmaximal ideal is the zero ideal and the Zariski tangent space is 0-dimensional. However,dimK T (X)x = 1 since X is nonsingular of dimension 1. Thus Θ(X)x 6= T (X)x . However, it is true that a nonsingular point is regular if we assume that k(x) is aseparable extension of k (see the proof below after (14.10)). Let us give another characterization of a regular local ring in terms of generatorsof its maximal ideal.

Lemma 14.7. (Nakayama). Let A be a local ring with maximal ideal m, and let Mbe a finitely generated A-module. Assume that M = N + mM for some submoduleN of M . Then M = N .

for some aij in m. Let R = (aij ) be the matrix of coefficients. Since (f1 , . . . , fr ) is asolution of the homogeneous system of equations R · x = 0, by Cramer’s rule,

det(R)fi = 0, i = 1, . . . , r. 135

However, det(R) = (−1)r + a for some a ∈ m (being the value of the characteristicpolynomial of R at 1). In particular det(R) is invertible in A. This implies that fi = 0for all i, i.e., M = {0}.

Corollary 14.8. 1. Let A be a local Noetherian ring and m be its maximal ideal.Elements a1 , . . . , ar generate m if and only if their residues modulo m2 span m/m2 asa vector space over k = A/m. In particular, the minimal number of generators of themaximal ideal m is equal to the dimension of the vector space m/m2 .

Proof. This follows from Proposition 14.4.

Definition 14.3. A system of parameters in a local ring A is a set of n = dim A

for some s > 0).

It follows from the proof of Proposition 14.4 that local rings OX,x always containa system of parameters. A local ring is regular, if and only if it admits a system ofparameters generating the maximal ideal. Such system of parameters is called a regularsystem of parameters. Let a1 , . . . , an be a system of parameters in OX,x , Choose an U be an open affineneighborhood of x such that a1 , . . . , an are represented by some regular functionsφ1 , . . . , φn on U . Then V (φ1 , . . . , φn ) ∩ U is equal to the closure of x in U corre-sponding to the prime ideal p ⊂ O(U ) such that OX,x ∼ = O(U )p ). In fact, the radicalof (φ1 , . . . , φn ) must be equal to p.

Example 14.10. 1. Let X be given by the equation Z12 + Z23 = 0 and x = (0, 0).The maximal ideal mX,x is generated by the residues of the two unknowns. It is easyto see that this ideal is not principal. The reason is clear: x is a singular point of Xand embdimx X = 2 > dimx X = 1. On the other hand, if we replace X by the setgiven by the equation Z12 + Z23 + Z2 = 0, then mX,x is principal. It is generated bythe germ of the function Z1 . Indeed, Z12 = −Z2 (Z22 + 1) and the germ of Z22 + 1 atthe origin is obviously invertible. Note that the maximal ideal m(X)x of O(X) is notprincipal.136 LECTURE 14. LOCAL PARAMETERS

codim(Θ(Y )x , Θ(X)x ) = dim Θ(X)x − dim Θ(Y )x ,

codimx (Y, X) = codimx X − codimx Y,

δx (Y, X) = codimx (Y, X) − codim(Θ(Y )x , Θ(X)x ). (14.6)

Proposition 14.11. Let Y be a closed subset of X and x ∈ Y . Assume x is a

regular point of X, then δx (Y, X) ≥ 0 and x is a regular point of Y if and only ifδx (Y, X) = 0.

In particular, x is a regular point of Y if and only if the cosets of the germs of

the functions defining X in an neighborhood of x modulo m2X,y span a linear subspaceof codimension equal to codimx X − codimx Y . Applying Nakayama’s Lemma, we seethat this is the same as saying that X can be locally defined by codimx X − codimx Yequations in an open neighborhood of x whose germs are linearly independent modulom2X,x .138 LECTURE 14. LOCAL PARAMETERS

For example, if Y is a hypersurface in X in a neighborhood of x, i.e. codimx Y =

Next we will show that every function from OX,x can be expanded into a formalpower series in a set of local parameters at x. Recall that the k-algebra of formal power series in n variables k[[Z]] = k[[Z1 , . . . , Zn ]]consists of all formal (infinite) expressions X P = ar Z r , r

(∗) f − P[r] (f1 , . . . , fn ) ∈ mr+1

the assertion of the theorem. First of all, we have to verify that this map is welldefined, i.e. property P (∗) determines P uniquely. Suppose there exists another formalpower series Q(Z) = j Qj such that

Lemma 14.14. (Artin-Rees). Let A be noetherian commutative ring and m be an

ideal in A. For any finitely generated A-module M and its submodule N , there existsan integer n0 such that

m(mn ∩ N ) = (mn+1 M ) ∩ N, n ≥ n0 . 141

We omit the proof which can be found in any text-book in commutative algebra.Definition 14.5. Let φ : OX,x → k[[Z1 , . . . , Zn ]] be the injective homomorphismconstructed in Theorem 14.1. The image φ(f ) of an element f ∈ OX,x is called theTaylor expansion of f at x with respect to the local parameters f1 , . . . , fn .Corollary 14.15. The local ring OX,x of a regular point does not have zero divisors.Proof. OX,x is isomorphic to a subring of the ring k[[Z]] which, as is easy to see, doesnot have zero divisors .Corollary 14.16. A regular point of an algebraic set X is contained in a uniqueirreducible component of X.Proof. This immediately follows from Corollary 14.3. Indeed, assume x ∈ Y1 ∩ Y2where Y1 and Y2 are irreducible components of X containing the point x. ReplacingX by an open affine neighborhood, we may find a regular function f1 vanishing on Y1but not vanishing on the whole Y2 . Similarly, we can find a function f2 vanishing onX \ Y1 and not vanishing on the whole Y1 . The product f = f1 f2 vanishes on thewhole X. Thus the germs of f1 and f2 are the zero divisors in OX,x . This contradictsthe previous corollary.Remark 14.17. Note the analogy with the usual Taylor expansion which we learnin Calculus. The local parameters are analogous to the differences ∆xi = xi − ai .The condition f − [P ]r (f1 , . . . , fn ) ∈ mr+1 X,x is the analog of the convergence: thedifference between the function and its truncated Taylor expansion vanishes at thepoint x = (a1 , . . . , an ) with larger and larger order. The previous theorem shows thata regular function on a nonsingular algebraic set is like an analytic function: its Taylorexpansion converges to the function. For every commutative ring A and its proper ideal I, one can define the I-adicformal completion of A as follows. Let pn,k : A/I n+1 → A/I k+1 be the canonicalhomomorphism of factor rings (n ≥ k). Set Y ÂI = {(. . . , ak , . . . , an . . .) ∈ (A/I r+1 ) : pn,k (an ) = ak for all n ≥ k}. r≥0

It is easy to see that ÂI is a commutative ring with respect to the addition andmultiplication defined coordinate-wise. We have a canonical homomorphism:

i : A → ÂI , a 7→ (a0 , a1 , . . . , an , . . .)

where an = residue of a modulo I n+1 . Note the analogy with the ring of p-adicnumbers which is nothing else as the formal completion of the local ring Z(p) ofrational numbers ab , p - b.142 LECTURE 14. LOCAL PARAMETERS

The formal I-adic completion Â is a completion in the sense of topology. One

makes A a topological ring (i.e. a topological space for which addition and multi-plication are continuous maps) by taking for a basis of topology the cosets a + I n .This topology is called the I-adic topology in A. One defines a Cauchy sequence asa sequence of elements an in A such that for any N ≥ 0 there exists n0 (N ) suchthat an − am ∈ I N for all n, m ≥ n0 (N ). Two Cauchy sequences {an } and {bn }are called equivalent if limn→∞ (an − bn ) = 0, that is, for any N > 0 there existsn0 (N ) such that an − bn ∈ I N for all n ≥ 0. An equivalence class of a Cauchysequence {an } defines an element of Â as follows. For every N ≥ 0 let αN be theimage of an in A/I N +1 for n ≥ n0 (N ). Obviously, the image of αN +1 in A/I N +1is equal to αN . Thus (α0 , α1 , . . . , αN , . . .) is an element from Â. Conversely, anyelement (α0 , α1 , . . . , αn , . . .) in Â defines an equivalence class of a Cauchy sequence,namely the equivalence class of {an }. Thus we see that Â is the usual completion ofA equipped with the I-adic topology. If A is a local ring with maximal ideal m, then Â denotes the formal completionof A with respect to the m-adic topology. Note that this topology is Hausdorff. Tosee this we have to show that for any a, b ∈ A, a 6= b, there exists n > 0 such thata + mn ∩ b + mn = ∅. this is equivalent to the existence of n > 0 such that a − b 6∈ mn .This will follow if we show that ∩n≥0 mn = {0}. But this follows immediately fromNakayama’s Lemma as we saw in the proof of Theorem 14.13. Since the topology isHausdorff, the canonical map from the space to its completion is injective. Thus weget A ,→ Â.Note that the ring Â is local. Its unique maximal ideal m̂ is equal to the closureof m in Â. It consists of elements (0, a1 , . . . , an , . . .). The quotient Â/m̂ is iso-morphic to A/m = κ. The canonical homomorphism ˆ(A) → Â/m̂ is of course(a0 , a1 , . . . , an , . . .) → a0 . The local ring Â is complete with respect to its m̂-topology. A fundamental resultin commutative algebra is the Cohen Structure Theorem which says that any completeNoetherian local ring (A, m) which contains a field is isomorphic to the quotient ringκ[[T1 , . . . , Tn ]], where κ is the residue field and n = dimκ m/m2 . This of course appliesto our situation when A = ÔX,x , where x is not necessary a rational point of X. Inparticular, when x is a regular point, we obtain

ÔX,x ∼ = k(x)[[T1 , . . . , Tn ]] (14.10)

which generalizes our Theorem 14.1.

Let us use the isomorphism (14.10) to show that a nonsingular point is regular ifassume that the extension k(x)/k is separable (i.e. can be obtained as a separablefinite extension of a purely transcendental extension of k). We only sketch a proof. 143

to the inclusion map of the ring into its completion. Note that for any local ring (A, m)which contains k, the canonical homomorphism of A-modules ρA : Derk (A, K) → Homk (m/m2 , K)is injective. In fact, if M is its kernel, then, for any δ ∈ M we have δ(m) = 0. Thisimplies that for any a ∈ m and any x ∈ A, we have 0 = δ(ax) = aδ(x) + xδ(a) =aδ(x). Thus aδ = 0. This shows that mM = 0, and by Nakayama’s lemma we getM = 0. Composing α with ρOX,x we obviously get ρÔX,x . Since the latter is injective,α is injective. Now we show that it is surjective. Let δ ∈ Derk (OX,x , K). Since itsrestriction to m2X,x is zero, we can define δ(a + m2X,x ) for any a ∈ OX,x . For anyx = (x0 , x1 , . . .) ∈ ÔX,x we set δ̃(x) = δ(x1 ). It is easy to see that this defines aderivation of ÔX,x /m̂2 such that ρ(δ̃) = δ. So, we obtain an isomorphism of K-vector spaces: Derk (ÔX,x , K) ∼ = Derk (OX,x , K).

By Cohen’s Theorem, ÔX,x ∼ = k(x)[[T1 , . . . , Tn ]], where the pre-image of the field ofconstant formal series is a subfield L of ÔX,x isomorphic to k(x) under the projectionto the residue field. It is clear that the pre-image of the maximal ideal (T1 , . . . , Tn )is the maximal ideal of OX,x . Let DerL (ÔX,x , K) be the subspace of Derk (ÔX,x , K)of derivation trivial on L. Using the same proof as in Lemma 13.3 of Lecture 13, weshow that DerL (ÔX,x , K) ∼ = Θ(X)x . Now we have an exact sequence, obtained byrestrictions of derivations to the subfield L: 0 → DerL (ÔX,x , K) → Derk (ÔX,x , K) → Derk (L, K). (14.11)It is easy to see that dimK Derk (L, K) = algdimk L = algdimk k(x). In fact, Derk (k(t1 , . . . , tr ), K) ∼ = Kr(each derivation is determined by its value on each ti ). Also each derivation can beuniquely extended to a separable extension. Thus exact sequence (14.11) gives dimK Derk (ÔX,x , K) = dimK Derk (ÔX,x , K) ≤ embdimx X + algdimk k(x).This implies that embdimx (X) = dim OX,x and hence OX,x is regular. Let (X, x) be a pair that consists of an algebraic set X and its point x ∈ X.Two such pairs are called locally isomorphic if the local rings OX,x and OY,y areisomorphic. They are called formally isomorphic if the completions of the local ringsare isomorphic. Thus any pair (X, x) where x is a nonsingular point of X is isomorphicto a pair (An (K), 0) where n = dimx X. Compare this with the definition of a smooth(or complex manifold).144 LECTURE 14. LOCAL PARAMETERS

Theorem 14.18. A regular local ring is a UFD (= factorial ring).

The proof of this non-trivial result can be found in Zariski-Samuel’s Commutative

Agebra, vol. II. See the sketch of this proof in Shafarevich’s book, Chapter II, §3. Ituses an embedding of a regular ring into the ring of formal power series.

Corollary 14.19. Let X be an algebraic set, x ∈ X be its regular point, and Y be a

closed subset of codimension 1 which contains x. Then there exists an open subset Ucontaining x such that Y ∩ U = V (f ) for some regular function on U .

Proof. Let V be an open affine open neighborhood of x, g ∈ I(Y ∩ V ), and let gx be

Here is the promised application.

Recall that a rational map f : X− → Y from an irreducible algebraic set X to analgebraic set Y is a regular map of an open subset of X. Two rational maps are saidto be equal if they coincide on an open subset of X. Replacing X and Y by openaffine subsets, we find ourselves in the affine situation of Lecture 4. We say that arational map f : X− → Y is defined at a point x ∈ X if it can be represented by aregular map defined on an open subset containing the point x. A point x where f isnot defined is called a indeterminacy point of f .

φ1 , . . . , φn+1 on D(φ). Since OX,x is factorial, we may cancel the germs (φi )x bytheir common divisor to assume that not all of them are divisible by the germ φx ofφ. The resulting functions define the same map to Pn (K). It is not defined at theset of common zeroes of the functions φi . Its intersection with Z cannot contain anyopen neighborhood of x, hence is a proper closed subset of Z. This shows that wecan extend f to a larger open subset contradicting the maximality of U .

Corollary 14.21. Any rational map of a nonsingular curve to a projective set is a

regular map. In particular, two nonsingular projective curves are birationally isomorphicif and only if they are isomorphic.

This corollary is of fundamental importance. Together with a theorem on resolution

of singularities of a projective curve it implies that the set of isomorphism classes offield extensions of k of transcendence degree 1 is in a bijective correspondence withthe set of isomorphism classes of nonsingular projective algebraic curves over k.

Lecture we shall address the following question: Given a projective algebraic k-set X,what is the minimal N such that X is isomorphic to a closed subset of PN k (K)? Weshall prove that N ≤ 2 dim X + 1. For simplicity we shall assume here that k = K.Thus all points are rational, the kernel of the evaluation maps is a maximal ideal, thetangent space is equal to the Zariski tangent space, a regular point is the same as anonsingular point.

if and only if it is bijective and for every point x ∈ X the differential map (df )x :T (X)x → T (Y )f (x) is injective.

Proof. To show that f is an isomorphism it suffices to find an open affine covering

of Y such that for any open affine subset V from this covering the homomorphism ofrings f ∗ : O(V ) → O(f −1 (V )) is an isomorphism. The inverse map will be definedby the maps of affine sets V → f −1 (V ) corresponding to the inverse homomorphisms(f ∗ )−1 : O(f −1 (V )) → O(V ). So we may assume that X and Y are affine and alsoirreducible. Let x ∈ X and y = f (x). Since f is bijective, f −1 (y) = {x}. The homomorphismf induces the homomorphism of local rings fy∗ : OY,x → OX,x . Let us show that it ∗

makes OX,x a finite OY,x -module. Let m ⊂ O(Y ) be the maximal ideal correspondingto the point y and let S = O(Y ) \ m. We know that OY,x = O(Y )S , and, sincefiniteness is preserved under localizations, O(X)f ∗ (S) is a finite OY,x -module. I claimthat O(X)f ∗ (S) = OX,x . Any element in OX,x is represented by a fraction α/β ∈Q(O(X)) where β(x) 6= 0. Since the map f is finite and bijective it induces a

Remark 15.2. The assumption of finiteness is essential. To see this let us take X to bethe union of two disjoint copies of affine line with the origin in the second copy deleted,and let Y = V (Z1 Z2 ) be the union of two coordinate lines in A2 (K). We map thefirst copy isomorphically onto the lines Z1 = 0 and map the second component of Xisomorphically onto the line Z2 = 0 with the origin deleted. It is easy to see that allthe assumptions of Theorem 15.1 are satisfied except the finiteness. Obviously, themap is not an isomorphism. 149

Definition 15.2. We say that a line ` in Pn (K) is tangent to an algebraic set X at a

point x ∈ X if T (`)x is contained in T (X)x (both are considered as linear subspacesof T (Pn (K))x ).

Let E be a linear subspace in Pn (K) defined by a linear subspace Ē of K n+1 . For

Now we see that a line E is tangent to X at the point x if and only if Ē is contained inthe space of solutions of (15.1) In particular we obtain that the union of lines tangentto X at the point x is the linear subspace of Pn (K) defined by the system of linearhomogeneous equations n X ∂Fi (x)Tj = 0, i = 1, . . . m. (15.2) ∂Tj j=0

It is called the embedded tangent space and is denoted by ET(X)x .

Lemma 15.3. Let X be a projective algebraic set in Pn (K), a ∈ Pn (K) ⊂ X, the

linear projection map pa : X → Pn−1 (K) is an embedding if and only if every line `in Pn (K) passing through the point a intersects X in at most one point and is nottangent to X at any point.

Proof. The induced map of projective sets f : X → Y = pa (X) is finite and bijective.By Theorem 15.1, it suffices to show that the tangent map (df )x is injective. Withoutloss of generality we may assume that a = (0, . . . , 0, 1) and the map pa is given byrestriction to X of the projection p : Pn (K) \ {a} → Pn−1 (K) is given by the formula:

its nonsingular point. Then ET(X)x is a projective subspace in Pn (K) of dimensionequal to d = dimx X.

Proof. We know that ET(X)x is the subspace of Pn (K) defined by equation(15.2). So

it remains only to compute the dimension of this subspace. Since x is a nonsingularpoint of X, the dimension of T (X)x is equal to d. Now the result follows fromcomparing equations (15.1) and (15.2). The first one defines the tangent space T (X)xand the second ET(X)x . The (linear) dimension of solutions of both is equal to

where < x, y > denotes the line spanned by the points x, y.

151

Let X be a closed subset of Pn (K). We set

SechX = p−1 12 (X × X \ ∆X ),

SecX = closure of SechX in Z.

The projection p12 and the projection p3 : Z → Pn (K) to the third factor define theregular maps p : SecX → X × X, q : SecX → Pn (K).For any (x, y) ∈ X × X \ ∆X the image of the fibre p−1 (x, y) under the map q isequal to the line < x, y >. Any such lines is called a secant of X. The union of allhonest secants of X is equal to the image of SechX under the map q. The closure ofthis union is equal to q(secX ). It is denoted by sec(X) and is called the secant varietyof X.

equations (1), where x is considered as a variable point in X. Consider the projectionof Z to the first factor. Its fibres are the embedded tangent spaces. Since X isnonsingular, all fibres are of dimension dim X. As in the case of the secant varietywe conclude that Z is irreducible and its dimension is equal to 2 dim X. Now theprojection of Z to Pn is a closed subset of dimension ≤ 2 dim X. It is equal to thetangential variety Tan(X).152 LECTURE 15. PROJECTIVE EMBEDDINGS

Now everything is ready to prove the following main result of this Lecture:

Theorem 15.7. Every nonsingular projective d-dimensional algebraic set X can be

is an isomorphism unless either x lies on a honest secant of X or in the tangential

variety of X. Since all honest secants are contained in the secant variety sec(X) ofX, and

dim sec(X) ≤ 2 dim X + 1 < n, dim Tan(X) ≤ 2 dim X < n,

we can always find a point a 6∈ X for which the map pa is an isomorphism. Continuingin this way, we prove the theorem.

Corollary 15.8. Every projective algebraic curve (resp. surface) is isomorphic to a

curve (resp. a surface) in P3 (K) (resp. P5 (K)).

Remark 15.9. The result stated in the Theorem is the best possible for projectivesets. For example, the affine algebraic curve: V (T12 + Fn (T2 )) = 0, where Fn is apolynomial of degree n > 4 without multiple roots, is not birationally isomorphic toany nonsingular plane projective algebraic curve. Unfortunately, we have no sufficienttools to prove this claim. Let me give one more unproven fact. To each nonsingularprojective curve X one may attach an integer g ≥ 0, called the genus of X. If K = Cis the field of complex numbers, the genus is equal to the genus of the Riemann surfaceassociated to X. Each compact Riemann surface is obtained in this way. Now for anyplane curve V (F ) ⊂ P2 (K) of degree n one computes the genus by the formula

(n − 1)(n − 2) g= . 2Since some values of g cannot be realized by this formula (for example g = 2, 4, 5) weobtain that not every nonsingular projective algebraic curve is isomorphic to a planecurve. Let sec(X) be the secant variety of X. We know that it is equal to the closure ofthe union sec(X)h of honest secant lines of X. A natural guess is that the comple-mentary set sec(X) \ sec(X)h consists of the union of tangent lines to X, or in otherwords to the tangential variety Tan(X) of X. This is true. 153

Proof. Since sec(X) is equal to the closure of an irreducible variety sec(X)h andTan(X) is closed, it is enough to prove that sec(X)h ∪ Tan(X) is a closed set. Let Z ne the closed subset of X × Pn (K) considered in the proof of Lemma 15.6.Its image under the projection to X is X, and its fibre over a point x is isomorphicto the embedded tangent space ET(X)x . Its image under the projection to Pn is thevariety Tan(X). We can view any point (x, y) = ((x0 , . . . , xn ), (y0 , . . . , yn )) ∈ ET(X)as a pair x + y ∈ K[]n+1 satisfying the equations Fi (T ) = 0. Note that for X = Pnwe have ET(X) = Pn ×Pn . Consider a closed subset Z of ET(X)×ET (X)×ETPn (K)defined by the equations

rank[x + y, x0 + y 0 , x00 + y 00 ] < 3, (15.3)

where the matrix is of size 3×(n+1) with entries in K[]. The equations are of coursethe 3 × 3-minors of the matrix. By Chevalley’s Theorem, the projection Z 0 of Z toET(X) × ET(X) is closed. Applying again this theorem, we obtain that the projectionof Z 0 to Pn is closed. Let us show that it is equal to Sech (X) ∪ Tan(X). It is clear that the image (x, x0 , x00 ) of z = (x + y, x0 + y 0 , x00 + y 0 ) in X × X × Xsatisfies rank[x, x0 , x00 ] < 3. This condition is equivalent to the following. For anysubset I of three elements from the set {0, . . . , n} let |xI + yI , x0I + yI0 , x00I + yI00 |be the corresponding minor. Then equation (15.3) is equivalent to the equations

P3 (K). Then X cannot be isomorphically projected into P2 (K) from a point outsideX. In particular any plane nonsingular projective curve of degree > 1 is linearly normal.

Proof. Applying Theorem 15.10 and Lemma 15.3, we have to show that sec(X) =P3 (K). Assume the contrary. Then sec(X) is an irreducible surface. For any x ∈X, sec(X) contains the union of lines joining x with some point y 6= x in X. Since Xis not a line, the union of lines < x, y >, y ∈ Y, y 6= x, is of dimension > 1 hence equalto Sec(X). Pick up three non-collinear points x, y, z ∈ X. Then Sec(X) contains theline < x, y >. Since each point of Sec(X) is on the line passing through z, we obtainthat each line < z, t >, t ∈< x, y > belongs to Sec(X). But the union of these linesis the plane spanned by x, y, z. Thus Sec(X) coincides with this plane. Since X isobviously contained in sec(X) this is absurd.

If d = 2, this gives that any surface of degree > 1 in P3 (K) is linearly normal.This bound is sharp. To show this let us consider the Veronese surface X = v2 (P2 (K)in P5 (K). Then we know that it is isomorphic to the set of symmetric 3 × 3-matricesof rank 1 up to proportionality. It is easy to see, by using linear algebra, that sec(X)is equal to the set of symmetric matrices of rank ≤ 2 up to proportionality. This isa cubic hypersurface in P3 (K) defined by the equation expressing the determinant ofsymmetric matrix. Thus we can isomorphically project X in P4 (K).Remark 15.14. According to a conjecture of R. Hartshorne, any non-degenerate non-singular closed subset X ⊂ Pn (K) of dimension d > 2n/3 is a complete intersection(i.e. can be given by n − d homogeneous equations).

Definition 15.4. A Severi variety is a nonsingular irreducible algebraic set X in Pn (K)

of dimension d = 2(n−2)/3 which is not contained in a hyperplane and with sec(X) 6=Pn (K). 155

The following result of F. Zak classifies Severi varieties in characteristic 0:

Theorem 15.15. Assume char(K) = 0. Each Severi variety is isomorphic to one of

the following four varieties:

• (n = 2) the Veronese surface v2 (P2 (K)) ⊂ P5 (K);

• (n = 4) the Segre variety s2,2 (P2 (K) × P2 (K)) ⊂ P8 (K);

• (n = 8) the Grassmann variety G(2, 6) ⊂ P14 (K) of lines in P5 (K);

• (n = 16) the E6 -variety X in P2 (K).

The last variety (it was initially missing in Zak’s classification and was added tothe list by R. Lazarsfeld) is defined as follows. Choose a bijection between the set of27 lines on a nonsingular cubic surface and variables T0 , . . . , T26 . For each triple oflines which span a tri-tangent plane form the corresponding monomial Ti Tj Tk . Let Fbe the sum of such 45 monomials. Its set of zeroes in P26 (K) is a cubic hypersurfaceY = V (F ). It is called the Cartan cubic. Then X is equal to the set of singularitiesof Y (it is the set of zeroes of 27 partial derivatives of F ) and Y equals sec(X).From the point of view of algebraic group theory, X = G/P , where G is a simplyconnected simple algebraic linear group of exceptional type E6 , and P its maximalparabolic subgroup corresponding to the dominant weight ω defined by the extremevertex of one of the long arms of the Dynkin diagram of the root system of G. Thespace P26 (K) is the projectivization of the representation of G with highest weight ω. We only check that all the four varieties from Theorem 15.15 are in fact Severivarieties. Recall that the Veronese surface can be described as the space of 3 × 3symmetric matrices of rank 1 (up to proportionality). Since a linear combination oftwo rank 1 matrices is a matrix of rank ≤ 2, we obtain that the secant variety iscontained in the cubic hypersurface in P5 defining matrices of rank ≤ 2. Its equationis the symmetric matrix determinant. It is easy to see that the determinant equationdefines an irreducible variety. Thus the dimension count gives that it coincides withthe determinant variety. Similarly, we see that the secant variety of the Segre varietycoincides with the determinant hypersurface of a general 3 × 3 matrix. The thirdvariety can be similarly described as the variety of skew-symmetric 6 × 6 matrices ofrank 2. Its secant variety is equal to the Pfaffian cubic hypersurface defining skew-symmetric matrices of rank < 6. Finally, the secant variety of the E6 -variety is equalto the Cartan cubic. Since each point of the Severi variety is a singular point of thecubic, the restriction of the cubic equation to a secant line has two multiple roots.This easily implies that the line is contained in the cubic. To show that the secantvariety coincides with the cartan cubic is more involved, One looks at the projectivelinear representation of the exceptional algebraic group G of type E6 in P26 defining156 LECTURE 15. PROJECTIVE EMBEDDINGS

the group G. One analyzes its orbits and shows that there are only three orbits: theE6 -variety X, the Cartan cubic with X deleted and P26 with Cartan cubic deleted.Since the secant variety is obviously invariant under the action of G, it must coincidewith the Cartan cubic. Note that in all four cases the secant variety is a cubic hypersurface and its set ofsingular points is equal to the Severi variety. In fact, the previous argument shows thatthe secant variety of the set of singular points of any cubic hypersurface is containedin the cubic. Thus Theorem 15.15 gives a classification of cubic hypersurfaces in Pnwhose set of singular points is a smooth variety of dimension 2(n − 2)/3. There is a beautiful uniform description of the four Severi varieties. Recall thata composition algebra is a finite-dimensional algebra A over a field K (not necessarycommutative or associative) such that there exists a non-degenerate quadratic formΦ : A → K such that for any x, y ∈ A

Φ(x · y) = Φ(x)Φ(y).

According to a classical theorem of A. Hurwitz there are four isomorphism classes

of composition algebras over a field K of characteristic 0: K, Co, Ha and Oc ofdimension 1, 2, 4 and 8, respectively. Here

where x̄ is defined as above for A = Ca and H, x̄ = x for A = K, and x̄ = (h̄, −h0 )

for any x = (h, h0 ) ∈ Oc. For example, if K = R, then Co ∼ = C (complex numbers), Ha ∼ = H (quaternions),Oc = O (octonions or Cayley numbers). For every composition algebra A we can consider the set H3 (A) of Hermitian3 × 3-matrices (aij ) with coefficients in A, where Hermitian means aij = āji . Itsdimension as a vector space over K equals 3 + 3r, where r = dimK A. There isa natural definition of the rank of a matrix from H3 (A). Now Theorem 15.15 saysthat the four Severi varieties are closed subsets of P3r+2 defined by rank 1 matrices inH3 (A). The corresponding secant variety is defined by the homogeneous cubic formrepresenting the “determinant” of the matrix. 157

Let us define Pn (A) for any composition algebra as An+1 \{0}/A∗ . Then one viewthe four Severi varieties as the “Veronese surfaces” corresponding to the projectiveplanes over the four composition algebra. As though it is not enough of these mysterious coincidences of the classifications,we add one more. Using the stereographic projection one can show that

P1 (R) = S 1 , P1 (C) = S 2 , P1 (H) = S 4 , P1 (O) = S 8 ,

where S k denote the unit sphere of dimension k. The canonical projection

which has a structure of a smooth bundle with fibres diffeomorphic to the sphereS r−1 = {x ∈ A∗ : x · x̄ = 1}. In this way we obtain 4 examples of a Hopf bundle: asmooth map of a sphere to a sphere which is a fibre bundle with fibres diffeomorphic toa sphere. According to a famous result of F. Adams, each Hopf bundle is diffeomorphicto one of the four examples coming from the composition algebras. Is there any direct relationship between Hopf bundles and Severi varieties?

Problems.1. Let X be a nonsingular closed subset of Pn (K). Show that the set J(X) of secantor tangent lines of X is a closed subset of the Grassmann variety G(2, n + 1). LetX = v3 (P1 (K)) be a twisted cubic in P3 (K). Show that J(X) is isomorphic toP2 (K).2. Find the equation of the tangential surface Tan(X) of the twisted cubic curve inP3 (K).3. Show that each Severi variety is equal to the set of singular points of its secantvariety. Find the equations of the tangential variety Tan(X).4. Assume that the secant variety sec(X) is not the whole space. Show that any Xis contained in the set of singular points of sec(X).5. Show that a line ` is tangent to an algebraic set X at a point x ∈ X if and only ifthe restriction to ` of any polynomial vanishing on X has the point x as its multipleroot.6*. Let X be a nonsingular irreducible projective curve in Pn (K). Show that theimage of the Gauss map g : X → G(2, n + 1) is birationally isomorphic to X unlessX is a line.158 LECTURE 15. PROJECTIVE EMBEDDINGSLecture 16

Blowing up and resolution of

singularities

Let us consider the projection map pa : Pn (K) \ {a} → Pn−1 (K). If n > 1 it isimpossible to extend it to the point a. However, we may try to find another projectiveset X which contains an open subset isomorphic to Pn (K) \ {a} such that the mappa extends to a regular map p̄a : X → Pn−1 (K). The easiest way to do it is toconsider the graph Γ ⊂ Pn (K) \ {a} × Pn−1 (K) of the map pa and take for X itsclosure in Pn (K)×Pn−1 (K). The second projection map X → Pn−1 (K) will solve ourproblem. It is easy to find the bi-homogeneous equations defining X. For simplicitywe may assume that a = (1, 0, . . . , 0) so that the map pa is given by the formula(x0 , x, . . . , xn ) → (x1 , . . . , xn ). Let Z0 , . . . , Zn be projective coordinates in Pn (K)and let T1 , . . . , Tn be projective coordinates in Pn−1 (K). Obviously, the graph Γ iscontained in the closed set X defined by the equations

Assume that ti = 1. Then the matrix of coefficients of the system of linear equations(16.2) contains n − 1 unit columns so that its rank is equal to n − 1. This showsthat the fibre q −1 (t) is isomorphic, under the first projection X → Pn (K), to the linespanned by the points (0, t1 , . . . , tn ) and (1, 0, . . . , 0). On the other hand the firstprojection is an isomorphism over Pn (K) \ {0}. Since X is irreducible (all fibres ofq are of the same dimension), we obtain that X is equal to the closure of Γ. Byplugging z1 = . . . zn in equations (16.2) we see that the fibre of p over the point

(ii) σ −1 (0, . . . , 0) ∼ = Pn−1 (K).We express this by saying that σ “ blows up” the origin. Of course if we take n = 1nothing happens. The algebraic set B is isomorphic to An (K). But if take n = 2,then B is equal to the closed subset of A2 (K) × P1 (K) defined by the equation

Z2 T0 − T1 Z1 = 0.

It is equal to the union of two affine algebraic sets V0 and V1 defined by the conditionT0 6= 0 and T1 6= 0, respectively. We have

V0 = V (Z2 − XZ1 ) ⊂ A2 (K) × P1 (K)0 , X = T1 /T0 ,

V1 = V (Z2 Y − Z1 ) ⊂ A2 (K) × P1 (K)1 , Y = T0 /T1 .

If L : Z2 − tZ1 = 0 is the line in A2 (K) through the origin “with slope” t, then thepre-image of this line under the projection σ : B → A2 (K) consists of the union oftwo curves, the fibre E ∼ = P1 (K) over the origin, and the curve L̄ isomorphic to Lunder σ. The curve L̄ intersects E at the point ((0, 0), (1, t)) ∈ V0 . The pre-imageof each line L with the equation tZ2 − Z1 consists of E and the curve intersectingE at the point ((0, 0), (t, 1)) ∈ V1 . Thus the points of E can be thought as the setof slopes of the lines through (0, 0). The ”infinite slope” corresponding to the lineZ1 = 0 is the point (0, 1) ∈ V1 ∩ E. Let I be an ideal in a commutative ring A. Each power I n of I is a A-module andI n I r ⊂ I n+r for every n, r ≥ 0. This shows that the multiplication maps I n × I r →I n+r define a ring structure on the direct sum of A-modules

A(I) = ⊕n≥0 I n .

Moreover, it makes this ring a graded algebra over A = A(I)0 = I 0 . Its homogeneouselements of degree n are elements of I n . Assume now that I is generated by a finite set f0 , . . . , fn of elements of A. Considerthe surjective homomorphism of graded A-algebras

If we additionally assume that A is a finitely generated algebra over a field k, we caninterpret Ker(φ) as the ideal defining a closed subset in the product X × Pnk where Xis an affine algebraic variety with O(X) ∼ = A. Let Y be the subvariety of X definedby the ideal I.

Definition 16.1. The subvariety of X × Pnk defined by the ideal Ker(φ) is denotedby BY (X) and is called the blow-up of X along Y . The morphism σ : BY (X) → Xdefined by the projection X × Pnk → X is called the monoidal transformation or theσ-process or the blowing up morphism along Y .

Let us fix an algebraically closed field K containing k and describe the algebraicset BY (X)(K) as a subset of X(K)×Pn (K). Let Ui = X ×(Pn (K))i and BY (X)i =BY (X) ∩ Ui . This is an affine algebraic k-set with

O(BY (X)i ) ∼ = O(X)[T0 /Ti . . . . , Tn /Ti ]/Ker(φ)i

where Ker(φ)i is obtained from the ideal Ker(φ) by dehomogenization with respectto the variable Ti . The fact that the isomorphism class of BY (X) is independent ofthe choice of generators f0 , . . . , fn follows from the following

Since (a0 , a1 ) is a regular sequence, this implies that G(a) ∈ (a0 ) for any a ∈ A. Fromthis we deduce that all coefficients of G(Z1 ) are divisible by a0 so that we can cancela0 in the previous equation. Proceeding in this way we find, by induction on r, thatF is divisible by L1 . Now assume n > 1 and consider the map φ0 as the composition map

isV1exact. The previous proposition asserts only that this complex is exact at the term An .

Proposition 16.8. Let X be an affine irreducible algebraic k-set, I be an ideal in

O(X) generated by a regular sequence (f0 , . . . , fn ), and let Y = V (I) be the set ofzeroes of this ideal. Let σ : BY (X) → X be the blow-up of X along Y . Then for anyx∈Y, σ −1 (x) ∼ = Pn (K).The pre-image of every irreducible component of Y is an irreducible subset of BY (X)of codimension 1.

Proof. By Proposition 16.6, Z = BY (X) is a closed subset of X × Pn (K) defined by

from the maximal ideal of a regular local ring A is a regular sequence if and only ifdim A/(a1 , . . . , an ) = dim A − n. By this result, the germs of f0 , . . . , fn in OX,xform a regular sequence. Then it is easy to see that their representatives in someO(U ) form a regular sequence.

Theorem 16.10. Let σ : BY (X) → X be the blow-up of a nonsingular irreducible

affine algebraic k-set X along a nonsingular closed subset Y . Then the following istrue

(i) σ is an isomorphism outside Y ;

(ii) BY (X) is nonsingular;

(iii) for any y ∈ Y, σ −1 (y) ∼

= Pn (K), where n = codimy (Y, X) − 1 = dim X − dimy Y − 1;

(iv) for any irreducible component Yi of Y , σ −1 (Yi ) is an irreducible subset of

codimension one.

Proof. Properties (i) and (iv) have been already verified. Property (iii) follows fromProposition 16.8 and Lemma 16.9. We include them only for completeness sake. Using(i), we have to verify the nonsingularity of BY (X) only at points x0 with σ(x0 ) =y ∈ Y . Replacing X by an open affine neighborhood U of y, we may assume thatY = V (I) where I is an ideal generated by a regular sequence f0 , . . . , fn . By Lemma16.3, σ −1 (U ) ∼ = BY ∩U (U ) so that we may assume X = U . By Proposition 16.6,BY (X) ⊂ X × Pnk (K) is given by the equations: fi Tj − fj Ti = 0, i, j = 0, . . . , n. Letp = (y, t) ∈ BY (X) where y ∈ Y, t = (t0 , . . . , tn ) ∈ Pn (K). We want to verify that itis a nonsingular point of BY (X). Without loss of generality we may assume that thepoint p lies in the open subset W = BY (X)0 where t0 6= 0. Since

We see that the submatrix J1 of J formed by the first N columns is obtained

from the Jacobian matrix of Y computed at the point y by applying elementary rowtransformations and when deleting the row corresponding to the polynomial F0 . SinceY is nonsingular at y, the rank of J1 is greater or equal than N − dimx Y − 1 =N − dim X + n. So rank J ≥ N + n − dim X = N + n − dim BY (X). This impliesthat BY (X) is nonsingular at the point (y, z). The pre-image E = σ −1 (Y ) of Y is called the exceptional divisor of the blowingup σ : BY (X) → X. The map σ “blows down” E of BY (X) to the closed subset Yof X of codimension n + 1. Lemma 16.3 allows us to ‘globalize’ the definition of the blow-up. Let X be anyquasi-projective algebraic set and Y be its closed subset. For every affine open setU ⊂ X, Y ∩ U is a closed subset of U and the blow-up BY ∩ U (U ) is defined. Itcan be shown that for any open affine cover {Ui }i∈I of X, the blowing-ups σi :BUi ∩Y (Ui ) → Ui and σj : BUj ∩Y (Uj ) → Uj can be “glued together” along theirisomorphic open subsets σi−1 (Ui ∩ Uj ) ∼ = σj−1 (Uj ∩ Uj ). Using more techniques onecan show that there exists a quasi-projective algebraic set BY (X) and a regular mapσ : BY (X) → X such that σ −1 (Ui ) ∼ = BUi ∩Y (Ui ) and, under this isomorphism, the −1restriction of σ to σ (Ui ) coincides with σi . The next fundamental results about blow-ups are stated without proof.Theorem 16.11. Let f : X− → Y be a rational map between two quasi-projectivealgebraic sets. There exists a closed subset Z of X and a regular map f 0 : BZ (X) → Ysuch that f 0 is equal to the composition of the rational map σ : BZ (X) → X and f . Although it sounds nice, the theorem gives very little. The structure of the blowing-up along an arbitrary closed subset is very complicated and hence this theorem giveslittle insight into the structure of any birational map. It is conjectured that everybirational map between two nonsingular algebraic sets is the composition of blow-upsalong nonsingular subsets and of their inverses. It is known for surfaces and, undersome restriction, for threefolds.168 LECTURE 16. BLOWING UP AND RESOLUTION OF SINGULARITIES

Definition 16.3. A birational regular map σ : X̄ → X of algebraic sets is said to be

a resolution of singularities of X if X̄ is nonsingular and σ is an isomorphism over anyopen set of X consisting of nonsingular points.

The next fundamental result of Heisuki Hironaka brought him the Fields Medal in1966:

Theorem 16.12. Let X be an irreducible algebraic set over an algebraically closed

field k of characteristic 0. There exists a sequence of monoidal transformations σi :Xi → Xi−1 , i = 1, . . . , n, along nonsingular closed subsets of Xi−1 contained in theset of singular points of Xi−1 , and such that the composition Xn → X0 = X is aresolution of singularities.

A most common method for define a resolution of singularities is to embed a

variety into a nonsingular one, blow up the latter and see what happens with theproper inverse transform of the subvariety (embedded resolution of singularities).

C1 ∩ σ −1 (0) = V (Z1 , t2 − 1) = {(0, 1), (0, −1)},

C2 ∩ σ −1 (0) = V (Z2 , t02 − 1) = {(0, 1), (0, −1)}.

σ −1 (0) ∩ C consists of two points. Moreover, it is easy to see that the curve Cintersects the exceptional divisor E = σ −1 (0) transversally at the two points. So thepicture is as follows:

The restriction σ : C → Y is a resolution of singularities of Y .

Example 16.14. 4 This time we take Y = V (Z12 − Z23 ). We leave to the readerto repeat everything we have done in Example 1 to verify that the proper transformσ̄ −1 (Y ) is nonsingular and is tangent to the exceptional divisor E at one point. So,the picture is like this170 LECTURE 16. BLOWING UP AND RESOLUTION OF SINGULARITIES

The equations of the proper inverse transform X1 are obtained by dropping the firstfactors. In each piece Vi the equations Zi = 0 define the intersection of the properinverse transform X1 of X with the exceptional divisor E1 ∼ = P2 (K). It is empty set inV1 , the affine line U = 0 in V2 and V3 . The fibre of the map X1 → X over the origin isR1 ∼ = P1 (K). It is easy to see (by differentiation) that V1 and V2 are nonsingular butV3 is singular at the point (U, V, Z3 ) = (0, 0, 0). Now let us start again. Replace X byV3 ∼= V (Z12 + Z23 Z3 + Z32 ) ⊂ P3 (K) and blow-up the origin. Then glue the blow-upwith V1 and V2 along V3 ∩ (V1 ∪ V2 ). We obtain that the proper inverse transform X2of X1 is covered by V1 , V2 as above and three more pieces

V4 : 1 + U 3 V Z12 + V 3 Z1 = 0

V5 : U 2 + Z22 V + V 2 = 0, V6 : U 2 + V 3 Z32 + 1 = 0.The fibre over the origin is the union of two curves R2 , R3 each isomorphic to P1 (K).The equation of R2 ∪ R3 in V5 is U 2 + V 2 = 0. The equation of R2 ∪ R3 in V3is U 2 + 1 = 0. Since R1 ∩ V3 was given by the equation Z3 = 0 and we used thesubstitution Z3 = V Z2 in V5 , we see that the pre-image of R1 intersects R1 and R2at their common point (U, V, Z2 ) = (0, 0, 0) in V5 . This point is the unique singularpoint of X2 . Let us blow-up the origin in V5 . We obtain X3 which is covered by opensets isomorphic to V1 , V2 , V4 , V6 and three more pieces:

The pre-image of the origin in X4 is a curve R6 ∼ = P1 (K). It is given by the ho-

mogeneous equation T02 + T1 T2 + T22 in homogeneous coordinates of the exceptionaldivisor of the blow-up (compare it with Example 16.15). The image of the curve R1intersects R6 at one point. So we get a resolution of singularities σ : X̄ = X4 → Xwith σ −1 equal to the union of six curves each isomorphic to projective line. Theyintersect each other according to the picture: Let Γ be the graph whose vertices correspond to irreducible components of σ −1 (0)and edges to intersection points of components. In this way we obtain the graph It is the Dynkin diagram of simple Lie algebra of type E6 . • • • • •

• Figure 16.2:

Exercises.1. Prove that BV (I) (X) is not affine unless I is (locally ) a principal ideal.2. Resolve the singularities of the curve xn + y r = 0, (n, r) = 1, by a sequenceof blow-ups in the ambient space. How many blow-ups do you need to resolve thesingularity?3. Resolve the singularity of the affine surface X : Z12 + Z23 + Z33 = 0 by a sequenceof blow-ups in the ambient space. Describe the exceptional curve of the resolutionf : X̄ → X.4. Describe A(I), where A = k[Z1 , Z2 , ], I = (Z1 , Z22 ). Find the closed subset BI (A)of A2 (K) × P1 (K) defined by the kernel of the homomorphism φ : A[T0 , T1 ] →A(I), T0 → Z1 , T2 → Z22 . Is it nonsingular?5*. Resolve the singularities of the affine surface X : Z12 + Z23 + Z35 = 0 by a sequenceof blow-ups in the ambient space. Show that one can find a resolution of singularitiesf : X̄ → X such that the graph of irreducible components of f −1 (0) is the Dynkindiagram of the root system of a simple Lie algebra of type E8 .6*. Resolve the singularities of the affine surface X : Z1 Z23 + Z13 + Z32 = 0 by asequence of blow-ups in the ambient space. Show that one can find a resolution ofsingularities f : X̄ → X such that the graph of irreducible components of f −1 (0) isthe Dynkin diagram of the root system of a simple Lie algebra of type E7 .7*. Resolve the singularities of the affine surface X : Z1 (Z22 + Z1n ) + Z32 = 0 by asequence of blow-ups in the ambient space. Show that one can find a resolution ofsingularities f : X̄ → X such that the graph of irreducible components of f −1 (0) isthe Dynkin diagram of the root system of a simple Lie algebra of type Dn . 173

8*. Resolve the singularities of the affine surface X : Z1 Z22 + Z3n+1 = 0 by a sequenceof blow-ups in the ambient space. Show that one can find a resolution of singularitiesf : X̄ → X such that the graph of irreducible components of f −1 (0) is the Dynkindiagram of the root system of a simple Lie algebra of type An .9*. Let f : P2 (K)− → P2 (K) be the rational map given by the formula T0 →T1 T2 , T1 → T2 T3 , T2 → T0 T1 . Show that there exist two birational regular mapsσ1 , σ2 : X → P2 (K) with f ◦ σ1 = σ2 such that the restriction of each σi overP2 (K)j , j = 0, 1, 2 is isomorphic to the blow-up along one point.174 LECTURE 16. BLOWING UP AND RESOLUTION OF SINGULARITIESLecture 17

Riemann-Roch Theorem

Let k be an arbitrary field and K be its algebraic closure. Let X be a projective varietyover k such that X(K) is a connected nonsingular curve. A divisor on X is an element of the free abelian group ZX generated by the setX(K) (i.e. a set of maps X(K) → Z with finite support). We can view a divisor asa formal sum X D= n(x)x, x∈X(K)

where x ∈ X, n(x) ∈ Z and n(x) = 0 for all x except finitely many. The group lawis of course defined coefficient-wisely. We denote the group of divisors by Div(X). A divisor D is called effective if all its coefficients are non-negative. Let Div(X)+be the semi-group of effective divisors. It defines a partial order on the group Div(X):

D ≥ D0 ⇐⇒ D − D0 ≥ 0.

Any divisor D can be written in a unique way as the difference of effective divisors

Recall that k(x) is the residue field of the local ring OX,x . If k = K, then k(x) = k. The local ring OX,x is a regular local ring of dimension 1. Its maximal ideal isgenerated by one element t. We call it a local parameter. For any nonzero a ∈ OX,x ,let νx (a) be the smallest r such that a ∈ mrX,x .

(i) νx (ab) = νx (f ) + νx (g);

(ii) νx (a + b) ≥ min{νx (a), νx (b)} if a + b 6= 0.

ab = tνx (a)+νx (b) a0 b0 , a + b = tνx (a) (a0 + tνx (b)−νx (b) b0 )

This proves (i),(ii). Note that we have the equality in (ii) when νx (a) 6= νx (b).

Let f ∈ R(X) be a nonzero rational function on X. For any open affine neigh-borhood U of a point x ∈ X, f can be represented by an element in Q(O(X)). SinceQ(O(X)) = Q(OX,x ), we can write f as a fraction a/b, where a, b ∈ OX,x . We set

νx (f ) = νx (a) − νx (b).

It follows from Lemma 17.1 (i), that this definition does not depend on the way wewrite f as a fraction a/b.

coset are called linearly equivalent. We write this D ∼ D0 .

By definition, L(D) contains the zero function f (thinking that νx (f ) = ∞ for allx ∈ X). It follows from Lemma 17.2 that L(D) is a vector space over k. TheRiemann-Roch formula is a formula for the dimension of the vector space L(D).Proposition 17.4. (i) L(D) is a finite-dimensional vector space over k;

where the equality is taken in the field of fractions K((T )) of K[[T ]]. We call theright-hans side, the Laurent series of f at x. Consider the linear map

L(D) → ⊕x∈X(K) T −n(x) K[[T ]]/K[[T ]] ∼

= ⊕x∈X(K) K n(x) ,178 LECTURE 17. RIEMANN-ROCH THEOREM

which assigns to f the collection of cosets of the Laurent series of f modulo k[[T ]].The kernel of this homomorphism consists of functions f such that νx (f ) ≥ 0 for allx ∈ X(K), i.e., regular function on X. Since X(K) is a connected projective set, anyregular function on X is a constant. This shows that L(D)⊗k K is a finite-dimensionalvector space over K. This easily implies that L(D) is a finite-dimensional vector spaceover k. (ii) Let g ∈ L(D + div(f )), then

It follows from the previous Proposition that dimk L(D) depends only on thedivisor class of D. Thus the function dim : Div(X) → Z, D 7→ dimk L(D) factorsthrough a function on Cl(X) which we will continue to denote by dim.

Theorem 17.5. (Riemann-Roch). There exists a unique divisor class KX on X such

that for any divisor class D

dimk L(D) = deg(D) + dimk L(KX − D) + 1 − g,

where g = dimk L(KX ) (called the genus of X),

Before we start proving this theorem, let us deduce some immediate corollaries. Taking D from KX , we obtain

P (T0 , T1 )/(T1 − a0 T0 ) · · · (T1 − xi T0 ),

sufficiently large, and applying the Corollary, we find that g = 0. The fact that deg(div(f )) = 0 is used for the proof of the Riemann-Roch formula.We begin with proving this result which we will need for the proof. Another proof ofthe formula, using the sheaf theory, does not depend on this result.Lemma 17.8. (Approximation lemma). Let x1 , . . . , xn ∈ X, φ1 , . . . , φn ∈ R(X), andN be a positive integer. There exists a rational function f ∈ R(X) such that

νx (f − φi ) > N, i = 1, . . . , n.

Proof. We may assume that X is a closed subset of Pn . Choose a hyperplane H which

does not contain any of the points xi . Then Pn \ H is affine, and U = X ∩ (Pn \ H)is a closed subset of Pn \ H. Thus U is an affine open subset of X containing thepoints xi . This allows us to assume that X is affine. Note that we can find a functiongi which vanishes at a point xi and has poles at the other points xj , j 6= i. Oneget such a function as the ratio of a function vanishing at xi but not at any xj andthe function which vanishes at all xj but not at xi . Let fi = 1/(1 + gim ). Thenfi − 1 = −gim /(1 + gim ) has zeroes at the points xj and has zero at xi . By taking mlarge enough, we may assume that νxi (fi − 1), νxj (fi − 1) are sufficiently large. Nowlet f = f1 φ1 + · · · + fn φn .It satisfies the assertion of the lemma. Indeed, we have

νx1 (c1s φ(1) (1)

On the other hand, νx1 (R.H.S.) can be made arbitrary large. This contradictionproves the assertion.

Let Λ be the direct product of the fraction fields R(X)x of the local rings OX,x ,where x ∈ X. By using the Taylor expansion we can embed each R(X)x in thefraction field K((T )) of K[[T ]]. Thus we may view Λ as the subring of the ring offunctions K((T ))X = Maps(X, K((T )).The elements of Λ will be denoted by (ξx )x . We consider the subring AX of Λ formedby (ξx )x such that ξx ∈ OX,xPexcept for finitely many x’s. Such elements are calledadeles. For each divisor D = n(x)x, we define the vector space over the field k:

Λ(D) = {(ξx )x ∈ Λ : νx (ξx ) ≥ −n(x)}.

where φx is the element of R(X)x represented by φ. Recall that the field of fractionsof OX,x is equal to the field R(X). Such adeles are called principal adeles. We willidentify the subring of principal adeles with R(X).

Proof. Let f : X → P1 be the regular map defined by φ. Then, as we saw in the

Corollary 17.15. Assume deg(D) < 0. Then L(D) = {0}.

Set r(D) = deg(D) − dim L(D).By Corollary 17.6, this number depends only on the linear equivalence class of D. Notethat, assuming the Riemann-Roch Theorem, we have r(D) = g −1−dim L(K −D) ≤g − 1. This shows that the function D 7→ r(D) is bounded on the set of divisors. Letus prove it.

Lemma 17.16. . The function D 7→ r(D) is bounded on the set Div(X).

Proof. As we have already observed, it suffices to prove the boundness of this functionon Cl(X). By Proposition 17.13 (iii), for any two divisors D0 , D with D0 ≥ D,

r(D0 ) ≤ r(mD + div(P (φ))) = r(mD)).

This proves the assertion.

185

Corollary 17.17. For any divisor D

dim A/(Λ(D) + R(X)) < ∞.

Proof. We know that

r(D0 ) − r(D) = dim(Λ(D0 ) + R(X)/Λ(D) + R(X))

is bounded on the set of pairs (D, D0 ) with D0 ≥ D. Since every adele ξ belongs tosome space Λ(D), the falsity of our assertion implies that we can make the spaces(Λ(D0 )+R(X)/Λ(D)+R(X)) of arbitrary dimension. This contradicts the boundnessof r(D0 ) − r(D).

dimk L(D) = deg(D) + dimk H(D) − g + 1. (17.1)

To prove the Riemann-Roch Theorem, it suffices to show that

dimk H(D) = dimk L(K − D).

To do this we need the notion of a differential of the field X.

A differential ω of R(X) is a linear function on A which vanish on some subspaceΛ(D) + R(X). A differential can be viewed as an element of the dual space H(D)∗for some divisor D. Note that the set Ω(X) of differentials is a vector space over the field R(X).Indeed, for any φ ∈ R(X) and ω ∈ Ω(X), we can define

div(φω) = div(ω) + div(φ) ≥ K − D.

is the maximal divisor D0 such that α vanishes on Λ(D0 ), we have K 0 ≥ K − D. By

Proposition 17.19, α = φω for some φ ∈ R(X). Hence

K 0 − K = div(α) − div(ω) = div(φ) ≥ −D.

showing that φ ∈ L(D). This defines a linear map

H(K − D)∗ → L(D), α → φ.

Obviously, this map is the inverse of the map c.

The number g = dimk L(K) is called the genus of X. It is easy to see by goingthrough the definitions that two isomorphic curves have the same genus. Now we will give some nice applications of the Riemann-Roch Theorem. We havealready deduced some corollaries from the RRT. We repeat them.

coordinates, we may assume that this line is the line at infinity V (T0 ). Let D = di=1 . PIt is clear that every rational function φ from the space L(nD), n ≥ 0, is regular onthe affine part U = X ∩ (P2 \ V (T0 )). A regular function on U is a n element of 189

1, φ1 , φ21 , φ31 , φ2 , φ22 , φ1 φ2 .

The number of them is 7, hence they must be linearly dependent in L(6 · x). Let

a0 + a1 φ1 + a2 φ21 + a3 φ31 + a4 φ2 + a5 φ22 + a6 φ1 φ2 .

with not all coefficients ai ∈ k equal to zero. I claim that a5 6= 0. Indeed, assume thata5 = 0. Since φ21 and φ32 are the only functions among the seven ones which has pole190 LECTURE 17. RIEMANN-ROCH THEOREM

of order 6 at x, the coefficient a3 must be also zero. Then φ1 φ2 is the only functionwith pole of order 5 at x. This implies that a6 = 0. Now φ21 is the only function withpole of order 4, so we must have a2 = 0. If a4 6= 0, then φ2 is a linear combination of1 and φ1 , and hence belongs to L(2 · x). This contradicts the choice φ2 . So, we geta0 + a1 φ1 = 0. This implies that a0 = a1 = 0. Consider the map f : X → P1 given by the function φ1 . Since φ2 satisfies anequation of degree 2 with coefficients from the field f ∗ (R(P1 )), we see that [R(X) :R(P1 )] = 2. Thus, adding φ2 to f ∗ (R(P1 )) we get R(X). Let

a0 + a1 Z1 + a2 Z12 + a3 Z13 + a4 Z2 + a5 Z22 + a6 Z1 Z2 = 0.

Since k(X) = k(Φ∗ (Z1 ), Φ∗ (Z2 )) we see that X is birationally isomorphic to theaffine curve V (F ). Note that a3 6= 0, since otherwise, after homogenizing, we get aconic which is isomorphic to P1 . So, homogenizing F we get a plane cubic curve withequation

F (T0 , T1 , T2 ) = T0 T22 + T13 + a1 T02 T1 + a0 T03 ,

or, after dehomogenizing,

Z22 + Z13 + a1 Z1 + a0 = 0.

It is called the Weierstrass equation. Since the curve is nonsingular, the cubic poly-nomial Z13 + a1 Z1 + a0 does not have multiple roots. This occurs if and only if itsdiscriminant ∆ = 4a31 + 27a20 6= 0. (17.4) 191

Problems1. Show that a regular map of nonsingular projective curves is always finite.2. Prove that for any nonsingular projective curve X of genus g there exists a regularmap f : X → P1 of degree (= [R(X) : f ∗ (R(P1 ))]) equal to g = 1.3. Show that any nonsingular projective curve X of genus 0 with X(k) = ∅ isisomorphic to a nonsingular conic on P2k [Hint: Use that dim L(−KX ) > 0 to find apoint x with deg(1 · x) = 2].4. Let X be a nonsingular plane cubic with X(k) 6= ∅. Fix a point x0 ∈ X(k).For any x, y ∈ X let x ⊕ y be the unique simple pole of a non-constant functionφ ∈ L(x + y − x0 ). show that x ⊕ y defines a group law on X. Let x0 = (0, 0, 1),where we assume that X is given by equation (3). Show that x0 is the inflection pointof X and the group law coincides with the group law on X considered in Lecture 6.5. Prove that two elliptic curves given by Weierstrass equations Z22 +Z32 +a1 Z1 +a0 = 0and Z22 + Z32 + b1 Z1 + b0 = 0 are isomorphic if and only if a31 /a20 = b31 /b20 .6. Let X be a nonsingular curve in P1 × P1 given by a bihomogeneous equation ofdegree (d1 , d2 ). Prove that its genus is equal to